Probability - Class 10 Maths - Extra Questions

There are three boys and two girls. A committee of two is to be formed. Find the probability of events that the committee contains one boy and one girl.

$$n(S) = { }^{5}C_{2}$$
$$\therefore n(S) = 10$$
$$B =$$ one boy and one girl
$$B = \left \{(b_{1}, G_{1}), (b_{2}, G_{1}), (b_{3}, G_{1})(b_{1}, G_{2}), (b_{2}, G_{2}), (b_{3}, G_{3})\right \}$$
$$\therefore n(B) = 6$$
$$\therefore P(B) = \dfrac {n(B)}{n(S)}$$
$$= \dfrac {6}{10} = \dfrac {3}{5}$$


There are three boys and two girls. A committee of two is to be formed. Find the probability of event that the committee contains at least one girl

Given 3 boys and 2 girls

Selecting two of the five $$(n(S)) = { }^5C_{2}$$

$$\therefore n(S) = \dfrac{5!}{2!\cdot 3!}=10$$

$$A =$$ at least one girl

$$A = \left \{(b_{1}, G_{1}), (b_{2}, G_{1}), (b_{3}, G_{1}) (b_{1}, G_{2}), (b_{2}, G_{2}), (b_{3}, G_{2}), (G_{1}, G_{2})\right \}$$

$$\therefore n(A) = 7$$

$$\therefore P(A) = \dfrac {n(A)}{n(S)} = \dfrac {7}{10}$$


A die is thrown 20 times. Getting a number greater than 4 is considered a success. Find the mean and variance of the number of successes.

Clearly, the distribution of the 'number of succeses' is a binomial distribution with n = 20 and,
$$p= Probability \ of \ getting \ a \ number \ greater \ than \ 4 = \dfrac{2}{6}=\dfrac{1}{3}$$
Therefore,
$$q=1-p=1-\dfrac{1}{3}=\dfrac{2}{3}$$

Mean = np and Variance = npq
$$Mean = 20 \times \dfrac{1}{2} = 6.66$$ and $$Variance =20 \times \dfrac{1}{3} \times \dfrac{2}{3}=4.44$$

Hence, Mean = 6.66 and Variance = 4.44


Write the sum of the probabilities of all the elementary events of an experiment.

The sum of the probabilities of all the elementary events of an experiment is one (1)


One card is drawn from a well-shuffled deck of $$52$$ cards. Calculate the probability that the card will 
not be an ace

favorable outcomes=$$52-4$$ (As, $$4$$ aces are present)
total outcomes$$=52$$
$$\begin{array}{c}\Pr  = \cfrac{{48}}{{52}}\\ = \cfrac{{12}}{{13}}\end{array}$$

Thus the probability will be $$\cfrac{{12}}{{13}}$$



Define an elementary events.

An event having only one outcomes of the experiment.


On throwing a dice once once, the probability of occurrence of an even number is $$ 1 / 2 $$

Even numbers on dice are total $$ 3 $$ i.e. $$ 2, 4  $$ and $$ 6 $$
Probability of occurrance of an even number = $$ \frac { No.\quad of\quad even\quad number\quad on\quad dice }{ Total\quad no.\quad on\quad dice } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 }  $$


If two dice are rolled 12 times, obtain the mean and the variance of the distribution of successes, if getting a total greater than 4 is considered a success.

Let X denote the number of successes in 12 trials. Then, X follows binomial distribution with parameters $$n=12$$
p = Probability of getting a total greater than 4 in a single throw of a pair of dice.
$$p=1-\dfrac{6}{36}=\dfrac{5}{6}$$

$$q=1-p=1-\dfrac{5}{6}=\dfrac{1}{6}$$

Therefore,
Mean = $$np = \dfrac{5}{6} \times 12 = 10$$
And, 
Variance = $$npq = 12 \times \dfrac{5}{6} \times \dfrac{1}{6} =\dfrac{5}{3}$$.


What is an elementary event ?

An event having only one outcome of the random experiment is called an elementary event.


$$10p + 10 = 100$$

We have,
$$10p+10=100$$

$$10p=100-10$$

$$10p=90$$

$$p=9$$

Hence, this is the answer.


Complete the following statements:
(i) Probability of an event E + Probability of the event not E $$=$$ ______.
(ii) The probability of an event that cannot happen is _____. Such an event is called _____.
(iii) The probability of an event that is certain to happen is _____. Such an event is called ______.
(iv) The sum of the probabilities of all the elementary events of an experiment is _____.
(v) The probability of an event is greater than or equal to _____ and less than or equal to ______.

(i) Probability of event $$E$$ and probability of event not $$E = P(E) + P(\bar E) = \textbf{1}$$
(ii) Probability of ab event that cannot happen is $$\textbf{zero (0)}$$, Such an event is called $$\textbf{Impossible Events}$$.
(iii) The probability of an event that is certain to happen is $$\textbf{1}$$. Such an event is called $$\textbf{sure or certain event}$$.
(iv) The sum of the probabilities of all the elementary events of an experiment is $$\displaystyle \sum_{i=1}^n P(E_i) = \textbf1$$.
(v) The probability of an event is greater than or equal to $$\textbf{zero(0), impossible event}$$ and less than or equal to $$\textbf{1, sure or certain event}$$.


Two dice are thrown at the same time and the product of number appearing on them ia noted. Find the probability that the product is a prime number 

Here,
maximum outcomes : $$36$$

Possible outcomes as product of prime numbers : $$(1,2), (2,1), (1,3), (3,1), (1,5), (5,1)$$

Therefore, 
$$P$$ (getting product a prime number) $$=\cfrac{{6}}{{36}}$$
$$=\cfrac{{1}}{{6}}$$


A card is drawn at random from a well shuffled pack of $$52$$ playing cards. Find the probability that the drawn card is neither a king nor a queen.

Total number of possible outcomes $$= 52$$
Let $$E$$ be the event that the drawn card is neither a king nor a queen.
Total number of kings and queens $$= 4 + 4 = 8$$
$$\therefore$$, there are $$52 - 8 = 44$$ cards that are neither king nor queen.
Total number of favourable outcomes $$= 44$$
$$\therefore$$ Required probability$$=P\left(E\right)=\dfrac{Favourable\,\, outcomes}{Total\,\,  number\,\,  of\,\,  outcomes}=\dfrac{44}{52}=\dfrac{11}{13}$$


$$\begin{array} { l } { \text { Given that the two numbers appearing on throwing two dice are different. Find } } \\ { \text { the probability of the event 'the sum of numbers on the dice is } 4 \text { '. } } \end{array}$$


Number of Possible outcomes $$^6P_2=\dfrac{6!}{4!}=30$$

Number of favourable outcomes $$\{1,3\},\{2,2\},\{3,1\}=3$$

Probabilty is given as $$\dfrac{3}{30}=\dfrac 1{10}$$



If two die are rolled 12 times, obtain the mean and the variance of the distribution of successes, if getting a total greater than 4 is considered a success.

Let X denote the number of successes in 12 trials. Then, X follows binomial distribution with parameters $$n=12$$
p = Probability of getting a total greater than 4 in a single throw of pair of dice.
$$p=1-\dfrac{6}{36}=\dfrac{5}{6}$$

$$q=1-p=1-\dfrac{5}{6}=\dfrac{1}{6}$$

Therefore,
$$Mean = np = \dfrac{5}{6} \times 12 = 10 $$

And, $$Variance = npq = 12 \times \dfrac{5}{6} \times \dfrac{1}{6}=\dfrac{5}{3}$$


When two dice are rolled :
Find probability of getting an odd total.

Possible outcomes when two dice are rolled:
$$\text{S}=\{(1,1),(1,2),(1,3),....(2,1),(2,2),(2,3),...(6,5)(6,6)\}$$
$$\therefore$$ the number of possible outcomes in the sample space is $$36$$.
The outcomes for the event that the total is odd:
$$\text{E}=\{(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(3,6),(4,1),(4,3),(4,5),(5,2),(5,4),$$$$(5,6),(6,1),(6,3),(6,5)\}$$
The number of favourable outcomes is $$18$$.
$$\therefore \text{P(E)}=\dfrac{18}{36}\\\\ \ \ \ \ \ \ \ \ \ =\dfrac{1}{2}$$


A die is thrown thrice. Find the probability of getting an odd number number at least once.


Let $$A_i$$ denotes the event of getting an odd number in $$i^{th}$$ throw, $$i=1, 2, 3$$. Then,

$$A_i=\{1,3,5\}\Rightarrow n(A)=3$$

$$P(A_i)=\dfrac {3}{6}=\dfrac {1}{2}$$, $$i=1, 2, 3$$.

$$p(\bar A_i)=1-p(A)=1-\dfrac{1}{2}$$
 
Required probability $$=P(A_1\cup A_2 \cup A_3)$$

$$=1-P(\bar A_1\cap \bar A_2 \cap \bar A_3)$$

$$=1-P(\bar {A_1})\ P(\bar {A_2})\ P(\bar {A_3})$$   [$$A_1,A_2,A_3$$ are independents events]

$$=1-\left (1-\dfrac {1}{2}\right)\left (1-\dfrac {1}{2}\right)\left (1-\dfrac {1}{2}\right)$$

$$=1-\dfrac{1}{8}$$ 

$$=\dfrac {7}{8}$$



In a simultaneous throw of a pair of dice, find the probability of getting:
$$8$$ as the sum


1872891_1673949_ans_54af4bdfcabc4c6fa4ae141e30e3ba1d.jpg


A letter of English alphabet is chosen at random. What is the probability that the chosen letter is a consonant. 

$$\textbf{Step-1: Find the vowels & consonants and then apply basic probability.}$$

                 $$\text{Total no. of English alphabets}$$ $$= 26$$

                 $$\text{Total no. of Vowels}$$ $$= A, E, I, O, U = 5$$

                 $$\text{Total no. of consonants}$$ $$= 26 - 5 = 21$$

                 $$\therefore $$ $$\text{n (chosen letter is a consonant)}$$ $$= 21 = n (E)$$

                 $$\text{n (sample space)}$$ $$= 26 = n(S)$$

                 $$\therefore$$ $$\text{Let}$$ $$E = $$ $$\text{chosen letter is a consonant event .}$$

                 $$\therefore P(E) = \dfrac{n(E)}{n (S)} = \dfrac{21}{26}$$
$$\textbf{Hence, the probability that the chosen letter is a consonant is }$$ $$\boldsymbol{\dfrac{21}{26}}$$


A bag contain $$4$$ red, $$5$$ black and $$6$$ white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is:
not black 

Number of red balls $$=4$$
Number of black balls $$=5$$
Number of white balls $$=6$$
Total number of balls $$=4+5+6=15$$
$$\therefore$$ the total number of cases $$=15$$
Since there are $$5$$ black balls, the number of favourable outcomes is $$5$$.
So, P(a black ball) $$=\dfrac{\text{Number of favourable cases}}{\text{Total number of cases}}=\dfrac{5}{15}=\dfrac{1}{3}$$

Hence, P(not a black ball) $$=1-\text{P(a black ball)}=1-\dfrac{1}{3}=\dfrac{3-1}{3}=\dfrac{2}{3}$$


List all events associated with the random experiment of tossing of two coins. How many of them are elementary events?

We know that the sample space is set of all possible outcomes of an experiment.
When two coins are tossed there are 4 possible outcomes $$HH,HT,TH,TT$$.

So, sample space is $$\left \{HH, TH, HT, TT\right \}$$.

Now, we also know that an event is an outcome or a collection of outcomes for a random experiment. 
Hence, the total possible events for a random experiment of tossing of two coins are $$15$$: 
$$\left \{HH\right \}$$, $$\left \{HT\right \}$$, $$\left \{TH\right \}$$, $$\left \{TT\right \}$$, $$\left \{HH,\ TT\right \}$$, $$\left \{HH,\ TH\right \}$$, $$\left \{HH,\ HT\right \}$$, $$\left \{HT,\ TH\right \}$$, $$\left \{HT,\ TT\right \}$$, $$\left \{TH,\ TT\right \}$$, $$\left \{HH, \ HT, \ TT\right \}$$, $$\left \{HH, \ TH, \ TT\right \}$$, $$\left \{HH, \ HT, \ TH\right \}$$, $$\left \{HT, \ TH, \ TT\right \}$$ ,$$\left \{ HH, \ HT, \ TH, \ TT\right \}$$

We also know that an elementary event is an event having a single outcome.
Hence, there are $$4$$ elementary events out of the total possible $$15$$ events. They are:
$$\left \{HH\right \}$$, $$\left \{HT\right \}$$, $$\left \{TH\right \}$$, $$\left \{TT\right \}$$


Three coins are tossed once. Describe the following events associated with this ransom experiment.
$$A=$$ Getting three heads, $$B=$$ Getting two heads and one tail,
$$C=$$ Getting three tails,  $$D=$$ Getting a head on the first coin.
Which events are elementary events 

$$S(A)=\left\{H, H, H\right\}$$

$$S(B)=\left\{\left\{T, H, H\right\}\ \ \left\{H, T, H\right\}\ \ \left\{H, H, T\right\}\right\}$$

$$S(C)=\left\{T, T, T\right\}$$

$$S(D)=\left\{\left\{H, H, H\right\}\ \ \left\{H, H, T\right\}\ \ \left\{H, T, H\right\}\ \ \left\{H, T, T\right\}\right\}$$

Elementary $$=A, C$$

1434907_1668097_ans_65f91b707c2e46beb69f1b9f3b8fe5aa.png


Find the probability of getting 4 when a die is thrown once.

$$\textbf{Step 1: Find the probability of getting 4.}$$
                $$\text{Possible outcomes on rolling the dice are 1,2,3,4,5 and 6.}$$
                $$\text{Total number of outcomes = 6}$$
                $$\text{Number of favorable outcome = 1}$$

                $$\text{Probability  }=\dfrac{Number\ of\  favorable\ outcome}{Total\ number\ of\  outcomes}=\dfrac{1}{6}$$

$$\textbf{Hence, the required probability is }\mathbf{\dfrac{1}{6}}.$$


A bag contains $$5$$ white and $$9$$ red balls. One ball is drawn at random from the bag. Find the probability of getting (a) a white ball, (b) a red ball.



2160678_1984809_ans_4a36e28ea8c54f6f92065bf25f3b357c.jpeg


Class 10 Maths Extra Questions