Probability — Page 2 | JEE Mathematics

Q11

In a binomial distribution

B\left( {n,p = {1 \over 4}} \right),

if the probability of at least one success is greater than or equal to

{9 \over {10}},

then

n

is greater than :

A

{1 \over {\log _{10}^4 + \log _{10}^3}}

B

{9 \over {\log _{10}^4 - \log _{10}^3}}

C

{4 \over {\log _{10}^4 - \log _{10}^3}}

D

{1 \over {\log _{10}^4 - \log _{10}^3}}

Correct Answer

Option D

Solution

Given that, no of trials = n Probability of success (p) =

{{1 \over 4}}

$\therefore$ Probability of no success = 1 -

{{1 \over 4}}

=

{{3 \over 4}}

As we know, probability of at least one success = 1 - probability of no success $\therefore$ According to the question, 1 - (Probability of no success in n trials) $\ge$

{9 \over {10}}

$\Rightarrow$ 1 - P(x = 0) $\ge$

{9 \over {10}}

$\Rightarrow$ 1 -

{}^n{C_0}

{\left( {{1 \over 4}} \right)^0}

{\left( {{3 \over 4}} \right)^n}

$\ge$

{9 \over {10}}

$\Rightarrow$ 1 -

{\left( {{3 \over 4}} \right)^n}

$\ge$

{9 \over {10}}

$\Rightarrow$

{\left( {{3 \over 4}} \right)^n}

$\le$

{1 \over {10}}

$\Rightarrow$

{\left( {{4 \over 3}} \right)^n}

$\ge$

10

[Taking log on both sides] $\Rightarrow$

n\left( {\log _{10}^4 - \log _{10}^3} \right)

$\ge$

{\log _{10}^{10}}

$\Rightarrow$

n\left( {\log _{10}^4 - \log _{10}^3} \right)

$\ge$

1

$\Rightarrow$

n

$\ge$

{1 \over {\left( {\log _{10}^4 - \log _{10}^3} \right)}}

$\therefore$ Option (D) is correct.

Q12

If

C

and

D

are two events such that

C \subset D

and

P\left( D \right) \ne 0,

then the correct statement among the following is :

A

P\left( {{C \over D}} \right)

\ge P\left( C \right)

B

P\left( {{C \over D}} \right)

< P\left( C \right)

C

P\left( {{C \over D}} \right)

= {{P\left( D \right)} \over {P\left( C \right)}}

D

P\left( {{C \over D}} \right)

= P\left( C \right)

Correct Answer

Option A

Solution

Given that

C \subset D

means

C

is present entirely inside

D

. Which is shown below.

P\left( {{C \over D}} \right)

=

{{P\left( {C \cap D} \right)} \over {P\left( D \right)}}

=

{{P\left( C \right)} \over {P\left( D \right)}}

As

C \cap D

means common part of events C and D which is equal to C.

0 \le P\left( D \right) \le 1

$\therefore$

{{P\left( C \right)} \over {P\left( D \right)}} \ge P\left( C \right)

Note: Here we are dividing with

{P\left( D \right)}

which is

\le 1

and

\ge 0

, as we know on dividing with a number n in the range

0 \le n \le 1

we get always more than or equal to the original number.

Q13

Three numbers are chosen at random without replacement from

\left\{ {1,2,3,..8} \right\}.

The probability that their minimum is

3,

given that their maximum is

6,

is :

A

{3 \over 8}

B

{1 \over 5}

C

{1 \over 4}

D

{2 \over 5}

Correct Answer

Option B

Solution

Given set S =

\left\{ {1,2,3,..8} \right\}

Choosing 3 numbers from 8 numbers can be done

{{}^8{C_3}}

ways. Choosing 3 numbers from 8 numbers while minimum no is 3 can be done

1 \times {}^5{C_2}

ways. $\therefore$ Probablity P(min = 3) =

{{1 \times {}^5{C_2}} \over {{}^8{C_3}}}\,

Choosing 3 numbers from 8 numbers while maximum no is 6 can be done

1 \times {}^5{C_2}

ways. $\therefore$ Probablity P(max = 6) =

{{1 \times {}^5{C_2}} \over {{}^8{C_3}}}\,

Choosing 3 numbers from 8 numbers while minimum number 3 and maximum no is 6 can be done

1 \times {}^2{C_1} \times 1

ways. $\therefore$

P\left( {\min = 3 \cap \max = 6} \right)

=

{{1 \times {}^2{C_1} \times 1} \over {{}^8{C_3}}}

The probability that their minimum is

3,

given that their maximum is

6,

is :

P\left( {{{\min = 3} \over {\max = 6}}} \right)

=

{{P\left( {\min = 3 \cap \max = 6} \right)} \over {P\left( {\max = 6} \right)}}

=

{{{{{}^2{C_1}} \over {{}^8{C_3}}}} \over {{{{}^5{C_2}} \over {{}^8{C_3}}}}}

=

{{{}^2{C_1}} \over {{}^5{C_2}}}

=

{{1 \over 5}}

Q14

A multiple choice examination has

5

questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get

4

or more correct answers just by guessing is :

A

{{17} \over {{3^5}}}

B

{{13} \over {{3^5}}}

C

{{11} \over {{3^5}}}

D

{{10} \over {{3^5}}}

Correct Answer

Option C

Solution

Each question has 3 alternative and exactly one is correct.

$\therefore$ Probability of giving correct answer P(correct) =

{1 \over 3}

$\therefore$ Probability of giving wrong answer P(wrong) =

{2 \over 3}

Here student give 4 or more correct answer.

$\therefore$ Student give either 4 correct answer or 5 correct answer.

Using Binomial Distribution, Required probability =

{}^5{C_4}{\left( {{1 \over 3}} \right)^4}{\left( {{2 \over 3}} \right)^1}

+

{}^5{C_5}{\left( {{1 \over 3}} \right)^5}{\left( {{2 \over 3}} \right)^0}

=

{{11} \over {{3^5}}}

Q15

If

12

different balls are to be placed in

3

identical boxes, then the probability that one of the boxes contains exactly

3

balls is :

A

220{\left( {{1 \over 3}} \right)^{12}}

B

22{\left( {{1 \over 3}} \right)^{11}}

C

{{55} \over 3}{\left( {{2 \over 3}} \right)^{11}}

D

55{\left( {{2 \over 3}} \right)^{10}}

Correct Answer

Option C

Solution

1st ball can go any of the 3 boxes.

So total choices for 1st ball = 3 2nd ball can also go any of the 3 boxes.

So total choices for 2nd ball = 3 . . . . 12th ball can go any of the 3 boxes.

So total choices for 12th ball = 3 Total choices for all 12 balls =

3 \times

3 \times

3 \times

.................12 times = 312. Now question says choose 3 balls from 12 balls. So no of ways =

{}^{12}{C_3}

ways. And then put it in a box. No of ways we can put =

{}^{12}{C_3} \times 1

ways.

Now we have 9 balls left and we have to put those 9 balls in the remaining 2 boxes.

Each ball can go to any of the 2 boxes, so for each ball there is 2 choices.

$\therefore$ Total ways for 9 balls = 29 $\therefore$ Total ways we can put those 12 balls in the boxes =

{}^{12}{C_3} \times 1 \times {2^9}

$\therefore$ Required probability =

{{{}^{12}{C_3} \times 1 \times {2^9}} \over {{3^{12}}}}

=

{{55} \over 3}{\left( {{2 \over 3}} \right)^{11}}

So option (C) is correct.

Q16

Consider

5

independent Bernoulli's trials each with probability of success

p.

If the probability of at least one failure is greater than or equal to

{{31} \over 32},

then

p

lies in the interval :

A

\left( {{3 \over 4},{{11} \over {12}}} \right]

B

\left[ {0,{1 \over 2}} \right]

C

\left( {{11 \over 12},1} \right]

D

\left( {{1 \over 2},{{3} \over {4}}} \right]

Correct Answer

Option B

Solution

Here is 5 trials.

So according to Bernoulli trial n = 5 P( at least one failure) = 1 - P( no failure) According to question, 1 - P( no failure)

\ge {{31} \over {32}}

$\Rightarrow$ 1 - P( x = 5 )

\ge {{31} \over {32}}

[Note: no failure = all success] $\Rightarrow$ 1 -

{}^5{C_5}\,{p^5}

\ge {{31} \over {32}}

$\Rightarrow$ 1 -

{p^5}

\ge {{31} \over {32}}

$\Rightarrow$

{p^5}

$\le$

{1 \over {32}}

$\Rightarrow$

p

$\le$

{1 \over {2}}

$\therefore$

p

\in

\left[ {0,{1 \over 2}} \right]

Note:

p

can not be less than 0 as probability is always $\ge$ 0 and $\le$ 1.

Q17

Let E and F be two independent events. The probability that both E and F happen is

{1 \over {12}}

and the probability that neither E nor F happens is

{1 \over {2}}

, then a value of

{{P\left( E \right)} \over {P\left( F \right)}}

is :

A

{4 \over 3}

B

{3 \over 2}

C

{1 \over 3}

D

{5 \over 12}

Correct Answer

Option A

Solution

Let P(E) = x and P(F) = y Now,

P(E \cap F) = {1 \over {12}}

\Rightarrow P(E)P(F) = {1 \over {12}}

\Rightarrow xy = {1 \over {12}}

Also,

P(E' \cap F') = {1 \over 2}

\Rightarrow (1 - P(E))(1 - P(F)) = {1 \over 2}

\Rightarrow (1 - x)(1 - y) = {1 \over 2}

\Rightarrow 1 - x - y + xy = {1 \over 2}

\Rightarrow x + y = 1 + xy - {1 \over 2}

\Rightarrow x + y = 1 + {1 \over {12}} - {1 \over 2}

\Rightarrow x + y = {1 \over 2} + {1 \over {12}} = {7 \over {12}}

........ (1) Now,

{(x - y)^2} = {(x + y)^2} - 4xy

\Rightarrow {(x - y)^2} = {{49} \over {144}} - {1 \over 3} \Rightarrow {{49 - 98} \over {144}} \Rightarrow {1 \over {144}}

\Rightarrow (x - y) = {1 \over {12}} \Rightarrow x - y = {1 \over 2}

........ (2) From Eqs. (1) and (2), we get

(x + y)(x - y) = {7 \over {12}} + {1 \over {12}}

\Rightarrow x = {4 \over {12}};y = {3 \over {12}}

\Rightarrow {x \over y} = {4 \over 3} = {{P(E)} \over {P(F)}}

Q18

From a group of 10 men and 5 women, four member committees are to be formed each of which must contain at least one woman. Then the probability for these committees to have more women than men, is :

A

{{21} \over {220}}

B

{{3} \over {11}}

C

{{1} \over {11}}

D

{{2} \over {23}}

Correct Answer

Option C

Solution

The number of ways to form a committee having at least one woman is

= {}^5{C_1} \times {}^{10}{C_3} + {}^5{C_2} \times {}^{10}{C_2} + {}^5{C_3} \times {}^{10}{C_1} + {}^5{C_4}

= {{5!} \over {4!}} \times {{10!} \over {7! \times 3!}} + {{5!} \over {2!3!}} \times {{10!} \over {8!2!}} + {{5!} \over {3!2!}} \times {{10!} \over {9!1!}} + {{5!} \over {4!}}

= 5 \times {{10 \times 9 \times 8} \over {3 \times 2}} + {{5 \times 4} \over {2 \times 1}} \times {{10 \times 9} \over 2} + {{5 \times 4} \over {2 \times 1}} \times 10 + 5

= 600 + 450 + 100 + 5 = 1155

The number of ways to form a committee having more women than men is

{}^5{C_2} \times {}^{10}{C_2} + {}^5{C_4} = {{5!} \over {6!2!}} \times {{10!} \over {9!}} + {{5!} \over {4!}}

= 10 \times 10 + 5 = 105

Therefore, the probability for the committees to have more women than men is

{{105} \over {1155}} = {1 \over {11}}

Q19

Three persons P, Q and R independently try to hit a target. I the probabilities of their hitting the target are

{3 \over 4},{1 \over 2}

and

{5 \over 8}

respectively, then the probability that the target is hit by P or Q but not by R is :

A

{{21} \over {64}}

B

{{9} \over {64}}

C

{{15} \over {64}}

D

{{39} \over {64}}

Correct Answer

Option A

Solution

We have the following probabilities: $\bullet$ The probability that the target is hit by the person P is

{3 \over 4}

. $\bullet$ The probability that the target is not hit by the person P is

1 - {3 \over 4} = {1 \over 4}

. $\bullet$ The probability that the target is hit by the person Q is

{1 \over 2}

. $\bullet$ The probability that the target is not hit by the person Q is

1 - {1 \over 2} = {1 \over 2}

. $\bullet$ The probability that the target is hit by the person R is

{5 \over 8}

. $\bullet$ The probability that the target is not hit by the person R is

1 - {5 \over 8} = {3 \over 8}

.

Here, we have used the fact that if the probability of occurrence of an event is p, then the probability of non-occurrence of an event is

q = 1 - p

.

Therefore, the probability that the target is hit by P or Q and not by R is (Probability that the target is hit by P and not by Q and R) + (Probability that the target is hit by Q and not by P and R) + (Probability that the target is hit by both P and Q and not by R)

= \left( {{3 \over 4}} \right)\left( {{1 \over 2}} \right)\left( {{3 \over 8}} \right) + \left( {{1 \over 4}} \right)\left( {{1 \over 2}} \right)\left( {{3 \over 8}} \right) + \left( {{3 \over 4}} \right)\left( {{1 \over 2}} \right)\left( {{3 \over 8}} \right)

= {9 \over {64}} + {3 \over {64}} + {9 \over {64}} = {{9 + 3 + 9} \over {64}} = {{21} \over {64}}

Q20

For three events A, B and C, P(Exactly one of A or B occurs) = P(Exactly one of B or C occurs) = P (Exactly one of C or A occurs) =

{1 \over 4}

and P(All the three events occur simultaneously) =

{1 \over {16}}

. Then the probability that at least one of the events occurs, is :

A

{7 \over {16}}

B

{7 \over {64}}

C

{3 \over {16}}

D

{7 \over {32}}

Correct Answer

Option A

Solution

Given, P (A $\cap$ B $\cap$ C) =

{1 \over {16}}

P (exactly one of A or B occurs) = P(A) + P (B) – 2P (A $\cap$ B) =

{1 \over 4}

.....(1) P (Exactly one of B or C occurs) = P(B) + P (C) – 2P (B $\cap$ C) =

{1 \over 4}

.....(2) P (Exactly one of C or A occurs) = P(C) + P(A) – 2P (C $\cap$ A) =

{1 \over 4}

.....(3) Adding (1), (2) and (3),we get 2[ P(A) + P(B) + P (C) - P (A $\cap$ B) - P (B $\cap$ C) - P (C $\cap$ A)] =

{3 \over 4}

$\Rightarrow$ P(A) + P(B) + P (C) - P (A $\cap$ B) - P (B $\cap$ C) - P (C $\cap$ A) =

{3 \over 8}

$\therefore$ P(atleast one event occurs) = P (A $\cup$ B $\cup$ C) = P(A) + P(B) + P (C) - P (A $\cap$ B) - P (B $\cap$ C) - P (C $\cap$ A) + P (A $\cap$ B $\cap$ C) =

{3 \over 8} + {1 \over {16}}

=

{7 \over {16}}