Probability - Class 11 Engineering Maths - Extra Questions

Let electricity contract be represented by E and plumbing contract be P

So $$P\left( E \right) = \dfrac 25$$

$$P\left( P \right) = 1 - \dfrac{4} {7} = \dfrac{3} { 7}$$

$$P\left( {E \cup P} \right) = \dfrac 23$$

NOW $$P\left( {E \cap P} \right) = P\left( E \right) + P\left( P \right) - P\left( {E \cup P} \right)$$

$$ = \dfrac{2} { 5} + \dfrac{3} { 7} - \dfrac{2} {3}$$

$$ = {\dfrac{17}{105}}$$

We have,

$$P\left( A \right)=0.3,P\left( B \right)=0.6$$

So,$$P\left( \overline{A} \right)=1-0.3=0.7$$

And$$P\left( \overline{B} \right)=1-0.6=0.4$$

Part (a)

P(A and not B)$$=P\left( A \right)\times P\left( \overline{B} \right)$$

P(A and not B)$$=0.3\times 0.4=0.12$$

Part (b)

P(neither A and nor B)$$=P\left( \overline{A} \right)\times P\left( \overline{B} \right)$$

P(A neither A and nor B)$$=0.7\times 0.4=0.28$$

Now when coin is tossed and a die is rolled simultaneously, $$(2\times6)$$  possible result =

$$(Head,1) , (Head,2) , (Head,3) , (Head,4) , (Head,5) , (Head,6) ,(Head,3) , (Tail,1) , (Tail,2) , (Tail,3) , \\ (Tail,4) , (Tail,5) , (Tail,6)$$

Let $$A=\left \{a_1,

a_2, ..., a_n\right \}$$. For each $$a_1\epsilon A (1\leq i\leq n)$$ we have

the following four cases:(i) $$a_1\epsilon P$$

and $$a_1\epsilon Q$$
(ii) $$a_1\not {\epsilon} P$$ and $$a_1\epsilon Q$$
(iii) $$a_1\epsilon P$$ and $$a_1\not {\epsilon}Q$$
(iv) $$a_1\not {\epsilon} P$$ and $$a_1\not {\epsilon} Q$$ Thus, the total number of ways of choosing P and Q is $$4^n$$.

A) Out of these four

choices (i) is not favourable for $$P\cap Q=\phi$$. Thus, probability that$$P\cap Q$$ is $$(3/4)^n$$.

B) $$P\cap Q$$ can
contain exactly one element in $$^nC_1(3^{n-1})=n(3^{n-1})$$ ways.

$$\therefore$$

Probability that $$P\cap Q$$ is singleton $$=(3^{n-1}/4^n$$

C) $$P\cup Q=A$$ in

$$3^n$$ ways.

$$\therefore$$

Probability $$P\cup Q=A$$ is $$(3/4)^n$$.

D)  $$|P|=|Q|$$ in the following ways.

$$(^nC_0)^2+(^nC_1)^2+....+(^nC_n)^2=^{2n}C_n$$.

$${\textbf{Step - 1: Assume two events and consider the condition}}{\text{.}}$$

                  $${\text{Suppose that , A is event of passing the first exam}}{\text{.}}$$

                  $${\text{and B is the event of passing second exam}}{\text{.}}$$

                  $${\text{Given that probability of randomly chosen student passing first exam is }}$$

                  $${\text{0.8 and passing second exam is 0}}{\text{.7}}{\text{.}}$$                

                  $${\text{so , p}}\left( {\text{A}} \right) = 0.8$$

                  $${\text{p}}\left( {\text{B}} \right) = 0.7$$

                  $${\text{Also given that probability of passing at least one of the exam is 0}}{\text{.95}}{\text{.}}$$

                  $${\text{so p}} \subset {\text{A}} \cup {\text{B = 0}}{\text{.95}}$$

$${\textbf{Step - 2: Find probability of passing both exam}}{\text{.}}$$

                   $${\text{So event of passing both is }}\left( {{\text{A}} \cap {\text{B}}} \right)$$

                   $${\text{So we have to find P}}\left( {{\text{A}} \cap {\text{B}}} \right)$$

                   $${\text{As per rule we know P}}\left( {{\text{A}} \cup {\text{B}}} \right) = {\text{P}}\left( {\text{A}} \right) + {\text{P}}\left( {\text{B}} \right) - {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)$$

                    $$ \Rightarrow {\text{P(A}} \cap {\text{B)}} = {\text{P(A)}} + {\text{P(B)}} - {\text{P(A}} \cup {\text{B)}}$$

                    $$ \Rightarrow {\text{P(A}} \cap {\text{B)}} = 0.8 + 0.7 - 0.95$$

                    $$ \Rightarrow {\text{P(A}} \cap {\text{B)}} = 0.55$$

$${\textbf{Hence, Probability of passing both exam is 0}}{\textbf{.55}}$$

Number of students opted mathematics $$=n(M)=30$$

Number of students opted Biology $$=n(B)=32$$

Number of students opted both Mathematics and Biology $$=n\left( M\cap B \right) =24$$

Number of students opted Mathematics or Biology $$=\left( M\cup B \right) $$

$$n\left( M\cup B \right) =n(M)+n(B)-n\left( M\cap B \right) \\ \Rightarrow n\left( M\cup B \right) =30+32-24=38$$

(i) Probability that student opted Mathematics or Biology $$=\dfrac { n\left( M\cup B \right)  }{ n } =\dfrac { 38 }{ 60 } =\dfrac { 19 }{ 30 } $$

(ii) Probability that student has opted neither mathematics nor Biology $$=1-P(M\cup B)=1-\dfrac { 19 }{ 30 } =\dfrac { 11 }{ 30 } $$

Number os students opted only Mathematics $$=n(M)-n\left( M\cap B \right) =30-24=8$$

(iii) Probability of selecting students who opted only mathematics $$=\dfrac { 8 }{ 60 } =\dfrac { 2 }{ 15 } $$

Let event $$A\rightarrow$$ She passes in Mathematics

Let event $$B\rightarrow$$ She passes in computer sciences

$$P(A)=\dfrac {4}{5}$$

$$P(A\cap B)=\dfrac{1}{2}$$

To calculate $$\rightarrow P\left(\dfrac{B}{A}\right)$$

Using Bayes theorem,

$$P(A\cap B)=P\left(\dfrac{B}{A}\right)\cdot P(A)$$

$$\Rightarrow P\left(\dfrac{B}{A}\right)=\dfrac{1/2}{4/5}=\dfrac{5}{8}$$

Assume that total number of marbles in the jar be $$x$$

Given that number of yellow marbles in a jar $$= 10$$

Given that the probability of selecting a green marble $$=\dfrac{1}{4}$$

Also, the probability of selecting a white marble $$=\dfrac{1}{3}$$

We know that the total probability $$= 1$$

Hence, Probability of selecting a yellow marble $$= 1 – ($$Probability of selecting a white marble $$+$$ Probability of selecting a green marble$$)$$

So the probability of selecting a yellow marble $$=1-(\dfrac{1}{4}+ \dfrac{1}{3})=1-\dfrac{7}{12}=\dfrac{5}{12}$$--------(i)

But the probability of selecting a yellow marble $$=\dfrac{No. of Yellow Marbles}{Total no. of Marbles} = \dfrac{10}{x}$$

∴$$\dfrac{5}{12} = \dfrac{10}{x}$$         [ from (i) ]

 ∴  $$x = 24$$

Hence the total number of marbles $$= 24$$

$${\textbf{Step -1: Given specific problem is solved independently.}}$$

                $${\text{Probability of solving a problem by A is}}$$ $${E_1}.$$

                $$\therefore p\left( {{E_1}} \right) = \dfrac{1}{2}$$

                $${\text{And probability of solving a problem by B is}}$$ $${E_2}.$$

                $$\therefore p\left( {{E_2}} \right) = \dfrac{1}{3}$$

                $${\text{By Rule of complementary events we have,}}$$

                $$p\left( {\overline {{E_1}} } \right) = 1 - \dfrac{1}{2} = \dfrac{1}{2}$$

                $$p\left( {\overline {{E_2}} } \right) = 1 - \dfrac{1}{3} = \dfrac{2}{3}$$

$${\textbf{Step -2: Find the probability that problem is solved.}}$$

                $$p\left( {{E_1} \cup {E_2}} \right) = 1 - P\left( {{{\overline E }_1}} \right)P\left( {{{\overline E }_2}} \right)$$

               $$ \Rightarrow p\left( {{E_1} \cup {E_2}} \right) = 1 - \dfrac{1}{2} \times \dfrac{2}{3}$$

               $$ \Rightarrow p\left( {{E_1} \cup {E_2}} \right) = \dfrac{2}{3}$$

$${\textbf{Step -3: Probability that exactly one of them solves the problem.}}$$

                $$p\left( {{\text{one of them is solved}}} \right)$$ $$ = p\left( {{E_1}} \right)p\left( {{{\overline E }_2}} \right) + p\left( {{E_2}} \right)p\left( {{{\overline E }_1}} \right)$$

               $$ = \dfrac{1}{2} \times \dfrac{2}{3} + \dfrac{1}{3} \times \dfrac{1}{2}$$

               $$ = \dfrac{1}{2}$$

$${\textbf{Hence the Probability that the problem is solved }}$$$${\textbf{is }}$$$$\mathbf{\dfrac{2}{3}.}$$ $${\textbf{ and Probability that exactly one of them can be solved is }}$$$$\mathbf{\dfrac{1}{2}.}$$

Probability of an event happening $$p=0.3$$

Probability of the event not happening $$=?$$

We know that, "Probability of an event happening $$+$$ Probability of the event not happening$$=1$$"

$$\therefore$$ Probability of the event not happening $$=1-$$Probability of an event happening                 

                                                                     $$=1-0.3$$

                                                                     $$=0.7$$

$$\therefore$$ Probability of the event not happening is $$0.7$$

We have,

Total number of cards $$=96$$

Number on the card is a perfect square between $$91$$ and $$99$$ $$=0$$

Then,

Probability of the card is a perfect square between $$91$$ and $$99=\dfrac{0}{96}=0$$

We have,

Probability of getting white ball is $$=\dfrac{5}{20}=\dfrac{1}{4}$$

Now,

Probability of given white balls is drawn $${{3}^{rd}}$$ times on $${{6}^{th}}$$ drawn =1.

Getting 2 white balls in 5 drawn $$\times $$ Probability of white balls in $${{6}^{th}}$$ terms.

$$ =\left( ^{5}{{C}_{2}} \right)\times {{\left( \dfrac{5}{20} \right)}^{5}}\times \dfrac{5}{20} $$

$$ =\dfrac{5!}{3!\times 2!}\times \dfrac{5}{20}\times \dfrac{5}{20}\times \dfrac{5}{20}\times \dfrac{5}{20}\times \dfrac{5}{20}\times \dfrac{5}{20}\times \dfrac{5}{20} $$

$$ =\dfrac{5\times 4}{2\times 1}\times \dfrac{15625}{64000000} $$

$$ =\dfrac{1562510}{64000000} $$

$$ =\dfrac{1}{2048} $$

Hence, this is the answer.

We have,

Total number of cards$$=15$$

Number of card is divisible by $$3$$ $$=\left( 3,6,9,12,15 \right)=5$$

Then,

Probability of card is divisible by $$3$$$$=\dfrac{5}{15}=\dfrac{1}{3}$$

Total days of leap year$$=365$$

Sunday in aleap year$$=53$$

Not contain sunday$$=365-53=312$$

So,

Probability$$=\dfrac{312}{365}$$

Hence, this is the answer.

We have,

Total number of cards $$=52$$

Number of outcomes of getting a black king $$=2$$

Therefore,

Probability $$P\left( A \right)=\dfrac{2}{52}=\dfrac{1}{26}$$

Hence, this is the answer.

We have,

Total number of cards$$=15$$

Prime cards$$=2,3,5,7,11,13$$

Numer of prime cards$$=6$$

Probability of Prime card$$=\dfrac{6}{15}=\dfrac{2}{5}$$

We have,

Total number of cards$$=96$$

Prime number is less than $$30$$$$=\left( 2,3,5,7,11,13,17,19,23,29 \right)$$

Then, Number of prime $$=10$$

Probability of prime numbers$$=\dfrac{10}{96}=\dfrac{5}{48}$$