MCQ Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 7

An experiment succeeds twice as often as it fails. Find the chance that in the next six trials, there shall be at least four successes.

0%
$\displaystyle \frac{233}{729}$
0%
$\displaystyle \frac{64}{729}$
0%
$\displaystyle \frac{496}{729}$
0%
$\displaystyle \frac{432}{729}$

Explanation

Let

$p$ denotes the probability of success and

$q$ denotes the probability of failure.
Then

$p+q=1$
Given

$p=2q$ .
Thus

$\displaystyle p=\frac{2}{3}, q=\frac{1}{3}.$
Hence probability of at least four successes in the next six trials is

$\displaystyle=^{6}C_{4}p^{4}q^{2}+^{6}C_{5}p^{5}q+^{6}C_{6}p^{6}$

$\displaystyle =15\left ( \frac{2}{3} \right )^{4}\left ( \frac{1}{3} \right )^{2}+6\left ( \frac{2}{3} \right )^{5}\left ( \frac{1}{3} \right )+\left ( \frac{2}{3} \right )^{6}$

$=\dfrac{426}{729}$

$8$ coins are tossed simultaneously. The probability of getting at least 6 heads is

0%
$57/64$
0%
$229/256$
0%
$7/64$
0%
$37/256$

The probability that a marksman will hit a target is given as 1/Then his probability at least one hit in 10 shots is.

0%
$\displaystyle 1-\left ( \frac{4}{5}\right)^{10}$
0%
$\displaystyle \frac{1}{5}$
0%
$\displaystyle 1-\frac{1}{5},$
0%
none of these.

Explanation

Here given

$n = 10, p = \cfrac{1}{5}$
Event of hitting the target is independent to each other.
Hence required probability is

$=1-P(X=0) = 1-^{10}C_0\left(\cfrac{4}{5}\right)^{10}=1-\left(\cfrac{4}{5}\right)^{10}$

A fair coin is tossed n times. if the probability that head occurs

$6$ times is equal to the probability that head occurs

$8$ times, then value of

$n$ is

0%
$24$
0%
$48$
0%
$14$
0%
$16$

Explanation

Let X be the number of times head occurs. Then X follows a binomial
distribution with parameter n and

$p=1/2$ .
We are given that,

$P(X=6)=P(X=8)$ .

$\Rightarrow ^nC_6\left (\dfrac {1}{2}\right )^6\left (\dfrac {1}{2}\right )^{n-6}=^nC_8\left (\dfrac {1}{2}\right )^8\left (\dfrac {1}{2}\right )^{n-8}$

$\Rightarrow ^nC_6\left (\dfrac {1}{2}\right )^n=^nC_8\left (\dfrac {1}{2}\right )^n$

$\Rightarrow ^nC_6=^nC_8=^nC_{n-8}\Rightarrow 6=n-8$ or

$n=14$

$X$ and

$Y$ are independent binomial variate

$B (5, 1/2)$ and

$B (7, 1/2)$ , then

$P(X+Y=3)$ is

0%
$55/1024$
0%
$55/4098$
0%
$55/2048$
0%
None of these

Explanation

Using

$X+Y\sim B (12, 1/2)$
Required probability

$P(X+Y=3)= ^{12}C_3\left(\cfrac{1 }{2}\right)^{12}=\cfrac{55}{1024}$

A random variable X follows binomial distribution with mean a and variance b. Then

0%
$a > b > 0$
0%
$\dfrac {a}{b} > 1$
0%
$\dfrac {a^2}{a-b}$ is an integer
0%
$\dfrac {a^2}{a+b}$ is an integer

Explanation

Suppose

$X=B(n, p)$ . Then

$a=np, b=npq$
As

$0 < q < 1$ , we get

$0 < npq < np$

$\displaystyle \Rightarrow a > b > 0\Rightarrow \frac {a}{b} > 1$
Also,

$\displaystyle \frac {a^2}{a-b}=\frac {n^2p^2}{np-npq}=\frac {np}{1-q}=n$ , which is an integer.

In a hurdle race, a runner has probability

$p$ of jumping over a specific hurdle. Given that in

$5$ trials, the runner succeeded

$3$ times, the conditional probability that the runner had succeeded in the first trail is

0%
$\displaystyle\frac { 3 }{ 5 }$
0%
$\displaystyle\frac { 2 }{ 5 }$
0%
$\displaystyle\frac { 1 }{ 5 }$
0%
$\displaystyle\frac { 4 }{ 5 }$

Explanation

Let

$A$ denote the event that the runner succeeds exactly

$3$ times out of five and

$B$ denote the event that the runner succeeds on the first trial.

$\displaystyle P\left( B|A \right) =\frac { P\left( B\cap A \right) }{ P\left( A \right) }$

But

$P\left( B\cap A \right) =P($ succeeding in the first trial and exactly once in two other trials

$)$

$\displaystyle=p\left( ^{ 4 }{ { C }_{ 2 } }.{ p }^{ 2 }{ \left( 1-p \right) }^{ 2 } \right) =6{ p }^{ 3 }{ \left( 1-p \right) }^{ 2 }$

and

$\displaystyle P\left( A \right) =^{ 5 }{ { C }_{ 3 } }{ p }^{ 3 }{ \left( 1-p \right) }^{ 2 }=10{ p }^{ 3 }{ \left( 1-p \right) }^{ 2 }$

Thus,

$\displaystyle P\left( B|A \right) =\frac { 6{ p }^{ 3 }{ \left( 1-p \right) }^{ 2 } }{ 10{ p }^{ 3 }{ \left( 1-p \right) }^{ 2 } } =\frac { 3 }{ 5 }$

A fair die is tossed eight times. Probability that on the eighth throw a third six is observed is

0%
$\displaystyle^{ 8 }{ { C }_{ 3 } }\frac { { 5 }^{ 8 } }{ { 6 }^{ 8 } }$
0%
$\displaystyle^{ 7 }{ { C }_{ 2 } }\frac { { 5 }^{ 5 } }{ { 6 }^{ 8 } }$
0%
$\displaystyle^{ 7 }{ { C }_{ 2 } }\frac { { 5 }^{ 5 } }{ { 6 }^{ 7 } }$
0%
None of these

Explanation

on the eight throw third six must appear, which implies that on first seven throws there must be two six.

probability of getting

$6$ is

$1/6$ and probability of not getting

$6$ is

$5/6$ .

we must get two six in seven throws,

so the probability is

$^{ 7 }{ { C }_{ 2 } }{ (1/6) }^{ 2 }{ (5/6) }^{ 5 }$ .

now prob that the eight throw is six is

$1/6$ .

so the final required probability is

$^{ 7 }{ { C }_{ 2 } }{ (1/6) }^{ 2 }{ (5/6) }^{ 5 }{1/6}$ .

Suppose

$X~B(n, p)$ and

$P(X=3)=P(X=5)$ . If

$p > \dfrac 12$ , then

0%
$n\leq 7$
0%
$n > 8$
0%
$n\geq 9$
0%
none of these

Explanation

$P(X=3)=P(X=5)$ .

$^nC_3p^3(1-p)^{n-3}=^nC_5p^5(1-p)^{n-5}$

$\displaystyle \Rightarrow \frac {(n-3)(n-4)}{20}=\frac {(1-p)^2}{p^2} < 1$ , since

$p> 1/2$

$\Rightarrow (n-3)(n-4)< 20\Rightarrow n\leq 7$ .

$15$ coins are tossed, then the probability of getting

$10$ heads will be

0%
$\dfrac {511}{32768}$
0%
$\dfrac {1001}{32768}$
0%
$\dfrac {3003}{32768}$
0%
$\dfrac {3005}{32768}$

Explanation

$\therefore$ Required probability

$\displaystyle =\ ^{15}C_{10}\left (\dfrac {1}{2}\right )^{10} \left (\dfrac {1}{2}\right )^{5}$

$\displaystyle = ^{15}C_{5} \dfrac {1}{2^{15}}$

$= \dfrac {15\times 14\times 13\times 12\times 11}{5\times 4\times 3\times 2\times 1} \times \dfrac {1}{2^{15}}$

$= \dfrac {3003}{32768}$

A bag contains three tickets numbered

$1, 2$ and

$3$ . A ticket is drawn at random and put back in the bag, and this is done four times. The probability that the sum of the numbers drawn is even is

0%
$\dfrac {40}{81}$
0%
$\dfrac {41}{81}$
0%
$\dfrac {14}{27}$
0%
$\dfrac {13}{81}$

Explanation

We make three cases, namely four odd numbers, two odd numbers and zero odd number; keeping in mind that the sum of four odd numbers & two odd numbers comes out to be even.

Let

$X=$ number of odd numbers chosen.

$P$ (sum even)

$=P(X=4)+P(X=2)+P(X=0)$

$=\left (\dfrac {2}{3}\right )^4+^4C_2\left (\dfrac {2}{3}\right )^2\left (\dfrac {1}{3}\right )^2+\left (\dfrac {1}{3}\right )^4$

$=\dfrac {41}{81}$

If X follows a binomial distribution with parameters

$n=100$ and

$p=1/3$ , find r for which

$P(X=r)$ is maximum

0%
$33$
0%
$22$
0%
$11$
0%
$15$

Explanation

$P(X=r)$ will be maximum when r is the mode. There are

$2$ cases for mode of binomial distribution.

Case

$1$ : If

$(n+1)p$ is an integer the binomial distribution is bimodal and the two modal values are

$(n+1)p$ and

$(n+1)p-1$

Case

$2$ : If

$(n+1)p$ is not an integer there exist unique modal value and its the integral part of

$(n+1)p$

The given binomial distribution has parameters

$n=100,p=\cfrac { 1 }{ 3 }$

We have

$(n+1)p=\cfrac { 101 }{ 3 } =33.67$

Hence unique mode is

$33$ the integral part of

$(n+1)p$

The probability that

$X=3$ equals

0%
$\displaystyle \frac {25}{216}$
0%
$\displaystyle \frac {25}{36}$
0%
$\displaystyle \frac {5}{36}$
0%
$\displaystyle \frac {125}{216}$

Explanation

Here

$p=1/6, q=5/6$
We have

$P(X=r)=pq^{r-1}$
Now,

$\displaystyle P(X=3)=\frac {1}{6}\left (\frac {5}{6}\right )^2=\frac {25}{216}$

The probability that a marksman will hit a target is given as

$\displaystyle \frac{1}{5}$ ,then his probability of at least one hit in 10 shots is

0%
$\displaystyle 1-\left ( \frac{4}{5} \right )^{10}$
0%
$\displaystyle \frac{1}{5^{10}}$
0%
$\displaystyle 1-\frac{1}{5^{10}}$
0%
None of these

Explanation

He will hit the target, $\displaystyle P(A)=\dfrac{1}{5}.$ He will not hit the target, $\displaystyle P\bar{(A)}=\dfrac{4}{5}.$ Probability that he
will not hit the target in 10 shoots is $\displaystyle \left ( \dfrac{4}{5} \right )^{10}$
So, at least once, target will be hit.Required probability $\displaystyle =1-\left ( \dfrac{4}{5} \right )^{10}$

If the mean of a binomial distribution is

$25$ , then find the interval in which the standard deviation lies.

0%
$[0,5)$
0%
$(0,5]$
0%
$[0,25)$
0%
$(0,25]$

Explanation

We have,

$np=25$ .

Now,

$0\le p<1$ and

$0\le q<1$

$\Rightarrow 0\le npq\le np\Rightarrow 0\le \sqrt { npq } \le \sqrt { np }$

$\Rightarrow 0\le$ S.D.

$\le 5$ .

But

$p\neq 0$ ,

therefore

$0\le$ S.D.

$<5$

$\Rightarrow$ S.D.

$\in [0,5)$

Find the probability of getting

$4$ exactly thrice in

$7$ throws of a die

0%
$^7C_3\left (\dfrac {1}{6}\right )^3\left (\dfrac {5}{6}\right )^4$
0%
$^7C_4\left (\dfrac {1}{6}\right )^4\left (\dfrac {5}{6}\right )^3$
0%
$^7C_4\left (\dfrac {1}{6}\right )^3\left (\dfrac {5}{6}\right )^4$
0%
$^7C_3\left (\dfrac {1}{6}\right )^4\left (\dfrac {5}{6}\right )^3$

Explanation

Here

$n=7$ , probability of getting

$4$ in one throw is

$p = \cfrac{1}{6} \Rightarrow q = 1-p=\cfrac{5}{6}$
Thus probability of getting

$4$ exactly thrice in

$7$ throw is

$=P(X=3)= ^7C_3 p^3 q^4 =^7C_3 \left(\cfrac{1}{6}\right)^3\left(\cfrac{5}{6}\right)^4=^7C_4 \left(\cfrac{1}{6}\right)^3\left(\cfrac{5}{6}\right)^4$

Numbers are selected at random, one at a time, from the two digit numbers

$00, 01, 02,....., 99$ with replacement. An event

$E$ occurs if & only if the product of the two digits of a selected number is

$18.$ If four numbers are selected, find the probability that the event

$E$ occurs at least

$3$ times.

0%
$\dfrac{97}{25^4}$
0%
$\dfrac{1}{25^4}$
0%
$\dfrac{95}{25^4}$
0%
$\dfrac{90}{25^4}$

Explanation

Out of the numbers,

$00,01,02,...,99$ , those numbers the product of whose digits is

$18$ are

$29,36,63,92$ .

$\displaystyle\therefore P(E)=\frac { 4 }{ 100 } =\frac { 1 }{ 25 }, P(\overline { E } )=\frac { 24 }{ 25 }$
The probability that in selecting four numbers that event

$E$ occurs at least three times

$=\displaystyle^{ 4 }{ { C }_{ 1 } }{ \left( \frac { 1 }{ 25 } \right) }^{ 3 }\frac { 24 }{ 25 } +^{ 4 }{ { C }_{ 4 } }{ \left( \frac { 1 }{ 25 } \right) }^{ 4 }$

$\displaystyle=\frac { 4\times 24 }{ { 25 }^{ 4 } } +\frac { 1 }{ { 25 }^{ 4 } } =\frac { 97 }{ { 25 }^{ 4 } }$

$\therefore k=97$ .

Suppose

$X$ follows a binomial distribution with parameters

$n$ and

$p$ , where

$0<p<1$ . If

$\displaystyle \frac { P\left( X=r \right) }{ P\left( X=n-r \right) }$ is independent of

$n$ and

$r$ , then

0%
$\displaystyle p=\frac { 1 }{ 2 }$
0%
$\displaystyle p=\frac { 1 }{ 3 }$
0%
$\displaystyle p=\frac { 1 }{ 4 }$
0%
None of these

Explanation

We have

$\displaystyle \frac { P\left( X=r \right) }{ P\left( X=n-r \right) } =\frac { _{ }^{ n }{ { C }_{ r } }.{ p }^{ r }.{ \left( 1-p \right) }^{ n-r } }{ _{ }^{ n }{ { C }_{ n-r }. }{ p }^{ n-r }.{ \left( 1-p \right) }^{ r } }$

$\displaystyle =\frac { { \left( 1-p \right) }^{ n-2r } }{ { p }^{ n-2r } } ={ \left( \frac { 1-p }{ p } \right) }^{ n-2r }={ \left( \frac { 1 }{ p } -1 \right) }^{ n-2r }$
and

$\displaystyle \frac { 1 }{ p } -1>0\therefore$ the ratio will be independent of

$n$ and

$r$ is

$\displaystyle \frac { 1 }{ p } -1=1$ or

$\displaystyle p=\frac { 1 }{ 2 }$

$n$ distinct objects are distributed randomly into

$n$ distinct boxed

0%
then the probability that no box is empty is $\displaystyle \frac{n!}{n^{n}}$
0%
then the probability that exactly one box is empty is $\displaystyle \frac{n!^{n}C_{2}}{n^{n}}$
0%
then the probability that a particular box get exactly $r$ objects is $\displaystyle \frac{^{n}C_{r}(n-1)^{n-r}}{n^{n}}$
0%
None of these

Explanation

No box empty the number of favourable ways

$=$

$\displaystyle n!$

Exactly one box empty then no of favourable ways

$=$

$\displaystyle^{n}C_{1}^{n-1}C_{1}^{n}C_{2} (n -2)!$

A particular box get exactly

$r$ objects

$=$

$\displaystyle ^{n}C_{r} ( n-1 )^{n-r}$

Anand plays with Karpov

$3$ games of chess. The probability that he wins a game is

$0.5$ , looses with probability

$0.3$ and ties with probability

$0.2$ . If he plays

$3$ games then find the probability that he wins atleast two games

0%
$1/2$
0%
$1/3$
0%
$1/4$
0%
$1/5$

Explanation

Let

$p$ denote the probability of winning a game.
Hence

$p=0.5$
Now let

$q$ denote the probability of not wining a game, ie either losing or a tie.
Hence

$q=0.3+0.2=0.5$

$=1-q$ .
The number of games (number of Bernoulli's trial) are 3.
Therefore the probability of winning atleast 2 games out of 3 is

$=\:^{3}C_{2}p^{2}.q+\:^{3}C_{3}p^{3}q^{0}$

$=3.(0.5)^{2}.(0.5)+1(0.5)^{3}$

$=(0.5)^{3}[3+1]$

$=\dfrac{1}{8}.4$

$=\dfrac{1}{2}$ .

A fair die is rolled

$(2n+1)$ times. The probability that the faces with even number show odd number of times is

0%
$\displaystyle \frac {3n+1}{4n+2}$
0%
$\displaystyle \frac {3}{4}$
0%
$\displaystyle \frac {2}{3}$
0%
$\displaystyle \frac {1}{2}$

Explanation

Probability of showing even numbers in a throw

$=\displaystyle\frac { 3 }{ 6 } =\frac { 1 }{ 2 }$

$\therefore$ Required probability

$\displaystyle=_{ }^{ 2n+1 }{ { C }_{ 1 } }.\frac { 1 }{ 2 } .{ \left( \frac { 1 }{ 2 } \right) }^{ 2n }+_{ }^{ 2n+1 }{ { C }_{ 3 } }.{ \left( \frac { 1 }{ 2 } \right) }^{ 3 }.{ \left( \frac { 1 }{ 2 } \right) }^{ 2n+2 }+...+_{ }^{ 2n+1 }{ { C }_{ 2n+1 } }.{ \left( \frac { 1 }{ 2 } \right) }^{ 2n+1 }$

$\displaystyle={ \left( \frac { 1 }{ 2 } \right) }^{ 2n+1 }\left[ _{ }^{ 2n+1 }{ { C }_{ 1 } }+^{ 2n+1 }{ { C }_{ 3 } }+^{ 2n+1 }{ { C }_{ 2n+1 } } \right]$

$\displaystyle=\frac { 1 }{ { 2 }^{ 2n+1 } } \times { 2 }^{ 2n+1-1 }=\frac { 1 }{ 2 }$

The value of P(2) in a Binomial distribution when

$\displaystyle P=\frac{1}{6}$ and n = 5 is

0%
$\displaystyle \dfrac{3125}{7776}$
0%
$\displaystyle \dfrac{250}{7776}$
0%
$\displaystyle \dfrac{1250}{7776}$
0%
$\displaystyle \dfrac{25}{7776}$

Explanation

For Binomial distribution,

$\displaystyle P(r)=^{n}C_{r}p^{r}q^{n-r}$
Here

$\displaystyle p=\dfrac{1}{6},q=\dfrac{5}{6}n=5,r=2.$
So,

$\displaystyle p(2)=^{5}C_{2}\left ( \dfrac{1}{6} \right )^{2}\left ( \dfrac{5}{6} \right )^{3}=\dfrac{1250}{7776}.$

A fair coin is tossed at a fixed number of times. If the probability of getting exactly

$3$ heads equals the probability of getting exactly

$5$ heads, then the probability of getting exactly one head is

0%
$\dfrac{1}{64}$
0%
$\dfrac{1}{32}$
0%
$\dfrac{1}{16}$
0%
$\dfrac{1}{8}$

Explanation

Let the coin be tossed $n$ times.

Let getting head is consider to be success.
$\therefore p=\dfrac{1}{2}, q=1-p=1-\dfrac{1}{2}=\dfrac{1}{2}$
It is given that, $p(X=3)=p(x=5)$

$\Rightarrow {}^nC_3\left(\dfrac{1}{2}\right)^3\left(\dfrac{1}{2}\right)^{n-3}={}^nC_5\left(\dfrac{1}{2}\right)^5\left(\dfrac{1}{2}\right)^{n-5}$
$\Rightarrow ^nC_3=^nC_5$

$\Rightarrow n=3+5$ $[\because ^nC_x=^nC_y\Rightarrow x+y=n]$
$\Rightarrow n=8$

Now, $P(x=1)=\ ^8C_1\left(\dfrac{1}{2}\right)^1\left(\dfrac{1}{2}\right)^{8-1}$

$=^8C_1\times \left(\dfrac{1}{2}\right)^8=\dfrac{1}{32}$

The mean number of heads in three tosses of a coin is

0%
$\displaystyle \frac{5}{2}$
0%
$\displaystyle \frac{1}{2}$
0%
$\displaystyle \frac{3}{2}$
0%
$\displaystyle \frac{1}{8}$

Explanation

We know that mean of binomial distibution is np here

$\displaystyle n=3,P=\dfrac{1}{2},.$ So, mean

$\displaystyle =3\times \dfrac{1}{2}=\dfrac{3}{2}.$

Given

$\displaystyle X\sim N\left( 10,36 \right)$ what is the variance of X?

0%
$10$
0%
$\displaystyle \sqrt { 10 }$
0%
$36$
0%
$6$

Explanation

Given distribution is in the form of poisson's distribution where

$k=10, \lambda = 36$

and in poisson's distribution

$\lambda$ is nothing but the variance

$\therefore$ Variance

$=36$

What is the possibility of getting at least 6 heads if eight coins are tossed simultaneously?

0%
$\dfrac{37}{256}$
0%
$\dfrac{25}{57}$
0%
$\dfrac{1}{13}$
0%
$\dfrac{5}{27}$

Explanation

We have, $p = \dfrac{1}{2}, q = \dfrac{1}{2}, n = 8$ Probability (at least 6 heads) $= {}^8C_6\times \left(\dfrac{1}{2}\right)^6\times \left(\dfrac{1}{2}\right)^2 + {}^8C_7 \times\left(\dfrac{1}{2}\right)^{7} \left(\dfrac{1}{2}\right)+{}^{8}C_{8}\left(\dfrac{1}{2}\right)^8$ $=\left(\dfrac{8\times7}{1\times2}+8+1\right) \left(\dfrac{1}{2}\right)^8$ $= 37 \times \dfrac{1}{256} = \dfrac{37}{256}$

The probability that an event does not happen in one trial is

$0.8$ . The probability that the event happens atmost once in three trials is

0%
$0.896$
0%
$0.791$
0%
$0.642$
0%
$0.592$

Explanation

$p = 0.2, q = 0.8, n = 3$
Atmost one success means either

$0 success$ or

$1 success$ . So, either it is

$3 failures$ or

$2 failures$ .
Hence, Probability of atmost one success is

$= P(3 failures)+P(2failures)$
using binomial probability distribution formula

$=^{3}C_0 \times0.8^3+ ^{3}C_1 \times 0.8^2 \times 0.2= .512+.384=.896$

Suppose

$X$ follows a binomial distribution with parameters

$n$ and

$p$ , where

$0 < p <1 ,\dfrac {P(X = r)}{P(X = n - r)}$ is independent of

$n$ for every

$r$ , then

$p =$

0%
$\dfrac {1}{2}$
0%
$\dfrac {1}{3}$
0%
$\dfrac {1}{4}$
0%
$\dfrac {1}{8}$

Explanation

$\dfrac{p(x=r)}{P(x=n-r)}=\dfrac{^nC_rP^rq^{n-r}}{^nC_{n-r}P^{n-r}Q^r}$

$=(\dfrac{q}{p})^{n-2r}$

As it is independent of

$n$ for every

$r$ so

$(\dfrac{q}{p})=1$

$\Rightarrow q=p$

and as

$q+p=1$

$\Rightarrow p=q=\dfrac{1}{2}$

Given

$\displaystyle X\sim B\left( n,p \right)$
If

$\displaystyle E\left( X \right) =6,var\left( X \right) =4.2$ then what is the number of trials?

0%
$21$
0%
$10$
0%
$20$
0%
$7$

Explanation

Given mean

$=np=6$

and Variance

$=npq=4.2$

$\Rightarrow q=0.7, p=0.3$

$\therefore n(0.3) = 6 \Rightarrow n=20$

$X$ is a poisson variate such that

$\alpha =P(X = 1) = P (X = 2)$ then

$P (X = 4) =$

0%
$2\alpha$
0%
$\dfrac {\alpha}{3}$
0%
$\alpha e^{-2}$
0%
$\alpha e^{2}$

Explanation

$P\left( X=K \right) =\dfrac { { \lambda }^{ K }{ e }^{ -\lambda } }{ K! }$

$P\left( X=1 \right) =P\left( X=2 \right)$

$\Rightarrow \dfrac { \lambda { e }^{ -\lambda } }{ 1! } =\dfrac { { \lambda }^{ 2 }{ e }^{ -\lambda } }{ 2! }$

$\Rightarrow \lambda =2$ So,

$\alpha =2{ e }^{ -2 }$

Now,

$P\left( X=4 \right) =\dfrac { { \lambda }^{ 4 }{ e }^{ -\lambda } }{ 4! }$

$=\dfrac { { 2 }^{ 4 }{ e }^{ -2 } }{ 4! }$

$=\dfrac{\alpha}{3}$

Two persons A and B are throwing an unbiased six faced die alternatively, with the condition that the person who throws

$3$ first wins the game. If A starts the game, the probabilities of A and B to win the same are respectively

0%
$\dfrac {6}{11}, \dfrac {5}{11}$
0%
$\dfrac {5}{11}, \dfrac {6}{11}$
0%
$\dfrac {8}{11}, \dfrac {3}{11}$
0%
$\dfrac {3}{11}, \dfrac {8}{11}$

Explanation

$p=\dfrac{1}{6}$ and

$q=\dfrac{5}{6}$
A wins if he gets

$3$ on his 1st turn or he gets

$3$ on his second turn but B doesn't get

$3$ on his first turn and so on..

$\therefore p(A) = p + pq^{2} + pq^{4} + ......$

$= p(1 + q^{2} + q^{4} + .....)$

$= \dfrac {p}{1 - q^{2}}=\dfrac {\dfrac{1}{6}}{1-(\dfrac {5}{6})^2}= \dfrac {6}{11}$

$\therefore p(B) = 1-p(A)= \dfrac{5}{11}$

$x$ follows the Binomial distribution with parameters

$n=6$ and

$p$ and

$9P(X=4) = P(X=2)$ , then

$p$ is

0%
$\dfrac {1}{4}$
0%
$\dfrac {1}{3}$
0%
$\dfrac {1}{2}$
0%
$\dfrac {2}{3}$

Explanation

Given, $9P(X=4)=P(X=2)$

$\therefore 9\ ^6C_4\ p^4(1-p)^2=\ ^6C_2\ p^2(1-p)^4$

$\Rightarrow \dfrac{(1-p)^2}{p^2}=9$ ............ $\because \ ^6C_2=^6C_4$

$\Rightarrow \dfrac{1-p}{p}=3$

$\Rightarrow p=\dfrac {1}{4}$

The probability that an individual suffers a bad reaction from an in injection is

$0.001$ . The probability that out of

$2000$ individuals exactly three will suffer bad reaction is

0%
$\dfrac{2}{e^2}$
0%
$\dfrac{2}{5e^2}$
0%
$\dfrac{8}{5e^2}$
0%
$\dfrac{4}{3e^3}$

The number of times a fair coin must be tossed so that the probability of getting at least one head is

$0.8$ is-

0%
$7$
0%
$6$
0%
$5$
0%
$3$

Explanation

Let p denote the probability that we will get the head, then

$p=\dfrac 12, q=1-p=\dfrac 12$

Suppose the coin is tossed n times. Let X denotes the number of times he get head in n-trials.

$\therefore P(X=r)=^nC_rp^rq^{n-r}\\\implies = ^nC_r \left(\dfrac 12\right)^r\left(\dfrac 12\right)^{n-r}=^nC_r\left(\dfrac 12\right)^n, r=0, 1, 2, ..., n$

Now,

$P(X\ge 1)>0.8\\\implies P(X=1)+P(X=2)+...+P(X=n)>\dfrac 8{10}\\\implies [P(X=0)+P(X=1)+...+P(X=n)]-P(X=0)>\dfrac 8{10}\\\implies 1-P(X=0)>\dfrac 8{10}\quad [\because \text{sum of probability is 1}]\\\implies P(X=0)<1-\dfrac 8{10}\\\implies P(X=0)<\dfrac 2{10}\\\implies ^nC_0\left(\dfrac 12\right)^n<\dfrac 15\\\implies \left(\dfrac 12\right)^n<\dfrac 15$

This is true only when the value of n is equal to or greater than

$3$

$\therefore ^nC_0\left(\dfrac 12\right)^n<\dfrac 15\implies n=3, 4, 5, ...$

Hence the coin must be tossed at least

$3$ times.

A die is thrown a fixed number of times. If probability of getting even number

$3$ times is same as the probability of getting even number

$4$ times, then probability of getting even number exactly once is

0%
$\dfrac{1}{4}$
0%
$\dfrac{3}{128}$
0%
$\dfrac{5}{64}$
0%
$\dfrac{7}{128}$

Explanation

Let the number of times a die is thrown be n.

Let the event getting an even number be the success

Then, the probability of this event is

$p =\dfrac 12$ .

Let X = number of times an even number is obtained.

Then, it is given that,

$P(X=3) = P(X=4)$

Using binomial distribution,

$P ( X=x) = ^nC_x p^x (1-p)^{n-x}$

Hence, we get,

$^nC_3 \left(\dfrac 12\right)^3 \left(\dfrac 12\right)^{n-3} = ^nC_4 \left(\dfrac 12\right)^4 \left (\dfrac 12\right)^{n-4}$

solving for n, we get,

$^nC_3 = ^nC_4$

which is

$\dfrac {n!}{3! (n-3)!} = \dfrac {n!}{ 4! (n-4)! }$

Thus,

$4 = n-3 \implies n=7$

Now, we want

$P(X=1) = ^7C_1 \left(\dfrac 12\right)^1 \left(\dfrac 12\right) ^6=7\times\left(\dfrac 12\right)^7$

Hence the required probability =

$\dfrac 7{128}$

A fair coin is tossed

$99$ times. Let X be the number of times heads occurs. Then

$P(X=r)$ is maximum when r is

0%
$49$
0%
$52$
0%
$51$
0%
None of these

Explanation

Notice that

$x$ has binomial distribution with

$n = 99, p = q = 0.5$ .

Therefore the probability,

$P(x=r)=^rC_np^rq^{n-r}$

Since

$p = q = 0.5$ we obtain that

$P(x=r)=^{r}C_{99}(0.5)^{r}(0.5)^{99-r}=^rC_{99}(o.5)^{99}$

whence the maximum corresponds to the points most closest to

$\dfrac n2$

Since n is odd, there are two such values:

$r_1=\dfrac {n-1}{2}=\dfrac {98}2=49, r_2=\dfrac {n+1}2=\dfrac {100}2=50$

For these numbers the function

$P(x = r)$ has maximum.

Out of

$800$ families with

$4$ children each, the expected percentage of families having almost

$2$ girls, is:

0%
$68.75 \%$
0%
$69.25 \%$
0%
$62.50 \%$
0%
None of these

Explanation

Given: 800 families having 4 children each.

To find: the expected percentage of families having almost 2 girls

Sol:

Let X be the number of boys in a family, If p is the probability that a child is a boy and if genders are independent between children, the probability that a family ends up having k girls is

$P(X=x)=^nC_kp^k(1−p)^{n-k}$

Here,

$n=4, p = \dfrac 12$

Now we need to find for almost 2 girls, i.e.,

$P(X=0)+P(X=1)+P(X=2)\\\implies ^4C_0\left(\dfrac 12\right)^0\left(\dfrac 12\right)^4+ ^4C_1\left(\dfrac 12\right)^1\left(\dfrac 12\right)^3+ ^4C_2\left(\dfrac 12\right)^2\left(\dfrac 12\right)^2\\\implies \dfrac 1{16}+4\times \dfrac 1{16}+6\times \dfrac 1{16}=\dfrac {11}{16}=0.6875$

Hence the expected percentage of families having almost 2 girls, is

$68.75\%$

For a poisson distribution with parameter

$\lambda = 0.25$ , the value of the

$2^{nd}$ moment about the origin is

0%
$0.25$
0%
$0.3125$
0%
$0.0625$
0%
$0.025$

Explanation

Second moment about the origin is

$\lambda +{ \lambda }^{ 2 } = 0.25+{(0.25)}^{2} = 0.3125$
Therefore the correct option is

$B$ .

If the mean and standard deviation of a binomial distribution are

$12$ and

$2$ respectively, then the value of its parameter

$p$ is

0%
$\cfrac{1}{2}$
0%
$\cfrac{1}{3}$
0%
$\cfrac{2}{3}$
0%
$\cfrac{1}{4}$

Explanation

Given mean

$=np=12$ .....

$(1)$

And we know that variance is square of standard deviation

so variance

$npq=2^2=4$ ....

$(2)$

Divide both the equations

$q=\dfrac{1}{3}$

$p=1-q=\dfrac{2}{3}$

The mean and the variance in a binomial distribution are found to be

$2$ and

$1$ respectively. The probability

$P(X = 0)$ is

0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{8}$
0%
$\dfrac{1}{16}$

Explanation

Mean of a binomial distribution

$=np=2$

Variance of a binomial distribution

$npq=1$

$\implies q=\cfrac { 1 }{ 2 }$

But

$p+q=1$

$\implies p=\dfrac { 1 }{ 2 }$

$\implies n=4$

Then,

$P(X=0)={}_{ 0 }^{ 4 }{ C }\ { (\cfrac { 1 }{ 2 } ) }^{ 4 }=\cfrac { 1 }{ 16 }$

In a Poisson distribution if

$P\left( X=2 \right) =P\left( X=3 \right)$ then the value of its parameter

$\lambda$ is

0%
$6$
0%
$2$
0%
$3$
0%
$0$

Explanation

According to Poisson distribution,

$P(x=k)=\dfrac{\lambda^{k}e^{-\lambda}}{k!}$ .
Hence

$P(x=2)=P(x=3)$ implies

$\dfrac{\lambda^{2}e^{-\lambda}}{2!}=\dfrac{\lambda^{3}e^{-\lambda}}{3!}$

$\dfrac{1}{2}=\dfrac{\lambda^{3-2}}{6}$

$\lambda=\dfrac{6}{2}=3$ .

$X$ is a Poisson's variate such that

$P(X=1)=3P(X=2)$ , then find the variance of

$X$ .

0%
$\cfrac 38$
0%
$\cfrac 13$
0%
$\cfrac 23$
0%
$\cfrac 54$

Explanation

Fact: Poisson distribution is

$P(X=r)=\dfrac{e^{-\lambda}\lambda ^r}{r!}$

Now given

$P(X=1)=3P(X=2)$

$\Rightarrow \dfrac{e^{-\lambda}\lambda}{1!}=3\cdot \dfrac{e^{-\lambda} \lambda^2}{2!}$

$\Rightarrow \lambda =\dfrac{2}{3}=$ Variance

If mean and variance of a Binomial variable X are 2 and 1 respectively, then the probability that X takes a value greater than 1 is

0%
$\dfrac{2}{3}$
0%
$\dfrac{4}{5}$
0%
$\dfrac{7}{8}$
0%
$\dfrac{11}{16}$

Explanation

Solution:

Given that:

$np=2,npq=1$

$\Longrightarrow2\times q=1\Longrightarrow q=\cfrac12$

$\Longrightarrow p+q=1\Longrightarrow p=\cfrac12$

$\Longrightarrow n\times\cfrac12\times\cfrac12=1\Longrightarrow n=4$

$\therefore P(X>1)={}^4C_2p^2q^2+{}^4C_3p^3q^1+{}^4C_{4}p^4q^0$

$=6\left(\cfrac12\right)^2\left(\cfrac12\right)^2$

$+4\left(\cfrac12\right)^3\left(\cfrac12\right)^1$

$+1\left(\cfrac12\right)^41$

$=\left(\cfrac12\right)^4(6+4+1)=\cfrac{11}{16}$

Hence, D is the correct option.

If the mean and variance of a binomial distribution are

$4$ and

$2$ , respectively. Then, the probability of atleast

$7$ successes is

0%
$\dfrac {3}{214}$
0%
$\dfrac {4}{173}$
0%
$\dfrac {9}{256}$
0%
$\dfrac {7}{231}$

Explanation

Here,

$mean = 4$ and

$variance = 2$

$\Rightarrow np = 4$ and

$npq = 2$
So,

$\dfrac {npq}{np} = \dfrac {2}{4}\Rightarrow q = \dfrac {1}{2}$
Then,

$p = 1 - q = 1 - \dfrac {1}{2} = \dfrac {1}{2}$

$Mean = np = 4$

$\Rightarrow n\times \dfrac {1}{2} = 4\Rightarrow n = 8$

$\therefore P(X = r) ={ }^{n}C_{r} p^{r} q^{n - r}$

$= { }^{8}C_{r}\left (\dfrac {1}{2}\right )^{8} \left [\because p = q = \dfrac {1}{2}\right ]$
The required probability of atleast

$7$ successes is

$P(X \geq 7) = P(X = 7) + P(X = 8)$

$= (^{8}C_{7} +{ }^{8}C_{8}) \left (\dfrac {1}{2}\right )^{8}$

$= \left (\dfrac {8!}{7!1!} + \dfrac {8!}{8!0!}\right )\left (\dfrac {1}{2}\right )^{8}$

$= (8 + 1)\left (\dfrac {1}{2}\right )^{8} = \dfrac {9}{256}$

Hence, option C is correct.

A box contains

$15$ green and

$10$ yellow balls. If

$10$ balls are randomly drawn, one-by-one, with replacement, then the variance of the number of green balls drawn is

0%
$\displaystyle\frac{12}{5}$
0%
$6$
0%
$4$
0%
$\displaystyle\frac{6}{25}$

Explanation

We can see that the Probability of getting a green ball is =

$\dfrac{\text{Favourable Events }}{\text{Total Number of balls}} = \dfrac{15}{25}$

Similarly, probability of getting a Yellow ball =

$\dfrac{10}{25}$

Let

$p = \dfrac{15}{25}$ and

$q = \dfrac{10}{25}$

As it follows binomial distribution,

$n= 10$

$Variance = npq = 10\times \dfrac{15}{25}\times\dfrac{10}{25}$

$= \dfrac{12}{5}$

Which of the following are correct regarding normal distribution curve?
(i) Symmetrical about the line

$X=\mu$ (Mean)
(ii) Mean

$=$ Median

$=$ Mode
(iii) Unimodal
(iv) Points of inflexion are at

$X=\mu \pm \sigma$

0%
(i), (ii)
0%
(ii), (iv)
0%
(i), (ii), (iii)
0%
All of these

Explanation

In a normal distribution , mean, median and mode are equal. The curve is bell type curve which is symmetric about

$x=$ mean (mode or median).
It is unimodal as it has its first derivative

$0$ at

$x=\mu$ (mean) and the derivative is less than

$0$ , for

$x>\mu$ and greater than

$0$ for

$x<\mu$
It has its point of inflection (second derivative

$0$ ) at

$x=\mu+\sigma$ and

$x=\mu-\sigma$ where

$\sigma$ is the standard deviation.

The probability that a student get success in a competition is

$\dfrac { 3 }{ 4 }$ . The probability that exactly

$2$ out of

$4$ students get success, is

0%
$\dfrac { 9 }{ 41 }$
0%
$\dfrac { 25 }{ 128 }$
0%
$\dfrac { 1 }{ 5 }$
0%
$\dfrac { 27 }{ 128 }$

Explanation

Given, probability of success,

$p=\dfrac { 3 }{ 4 }$ and probability of failure,

$q=1-p=\dfrac { 1 }{ 4 }$
Here, total no. of students is

$n=4$
Using binomial distribution,
Required probability

$={}^{n}C_{r}\cdot p^{n-r}\cdot q^{r}$

$={}^{ 4 }{ C }_{ 2 }{ \left( \dfrac { 3 }{ 4 } \right) }^{ 2 }{ \left( \dfrac { 1 }{ 4 } \right) }^{ 2 }$

$=\dfrac { 4! }{ 2!2! } \times \dfrac { 9 }{ 16 } \times \dfrac { 1 }{ 16 } =\dfrac { 27 }{ 128 }$

The mode of the binomial distribution for which mean and standard deviation are

$10$ and

$\sqrt { 5 }$ respectively, is

0%
$7$
0%
$8$
0%
$9$
0%
$10$

Explanation

In a binomial distribution, mean,

$np=10$ and variance,

$npq=5$

[Since, variance=square of standard deviation]

$\therefore 10\times q=5\Rightarrow q=\dfrac { 1 }{ 2 }$

$\Rightarrow p=1-q=1-\dfrac { 1 }{ 2 } =\dfrac { 1 }{ 2 }$
Let

$x$ be the mode. Then,

$np+q > x > np-q$

$\Rightarrow 10+\dfrac { 1 }{ 2 } > x > 10-\dfrac { 1 }{ 2 }$

$\Rightarrow \dfrac { 21 }{ 2 } > x > \dfrac { 19 }{ 2 } \Rightarrow 9.5 < x < 10.5$
Hence, option (d) is correct.

If r.v.

$X\sim B\left( n=5,P=\cfrac { 1 }{ 3 } \right)$ , then

$P(2< X< 4)=$ ...........

0%
$\cfrac { 80 }{ 243 }$
0%
$\cfrac { 40 }{ 243 }$
0%
$\cfrac { 40 }{ 343 }$
0%
$\cfrac { 80 }{ 343 }$

Explanation

$X$ can take only integer values, so

$P(2<X<4)=P(3)$ .

Also, let

$Q=1-P=\cfrac{2}{3}$

we have,

$P(3)=$

$^{ 5}C_{ 3 }.\left(\dfrac { 1 }{ 3 } \right)^{ 3 }.\left(\dfrac { 2 }{ 3 } \right)^{ 2 }$ ...... Using binomial distribution

$=\dfrac { 40 }{ 243 }$

answer (b).

The probability that an event does not happen in one trial is 0.8.The probability that the event happens atmost once in three trails is

0%
0.896
0%
0.791
0%
0.642
0%
0.592

CBSE Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 7 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 7 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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