Probability — Page 3 | JEE Mathematics

Q21

A box 'A' contains

2

white,

3

red and

2

black balls. Another box 'B' contains

4

white,

2

red and

3

black balls. If two balls are drawn at random, without eplacement, from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from box 'B' is :

A

{9 \over {16}}

B

{7 \over {16}}

C

{9 \over {32}}

D

{7 \over {8}}

Correct Answer

Option B

Solution

Probability of drawing a white ball and then a red ball from bag B is given by

{{{}^4{C_1} \times {}^2{C_1}} \over {{}^9{C_2}}}

=

{2 \over 9}

Probability of drawing a white ball and then a red ball from bag A is given by

{{{}^2{C_1} \times {}^3{C_1}} \over {{}^7{C_2}}}

=

{2 \over 7}

Hence, the probability of drawing a white ball and then a red ball from bag B =

{{{2 \over 9}} \over {{2 \over 7} + {2 \over 9}}}

=

{{2 \times 7} \over {18 + 14}}

=

{7 \over {16}}

Q22

If two different numbers are taken from the set {0, 1, 2, 3, ........, 10}; then the probability that their sum as well as absolute difference are both multiple of 4, is :

A

{{12} \over {55}}

B

{{14} \over {45}}

C

{{7} \over {55}}

D

{{6} \over {55}}

Correct Answer

Option D

Solution

Let A = {0, 1, 2, 3, 4, ......., 10} Total number of ways of selecting 2 different numbers from A is n (S) = 11C2 = 55, where 'S' denotes sample space Let E be the given event $\therefore$ E = {(0, 4), (0, 8), (2, 6), (2, 10), (4, 8), (6, 10)} $\Rightarrow$ n (E) = 6 $\therefore$ P(E) =

{{n\left( E \right)} \over {n\left( S \right)}}

=

{6 \over {55}}

Q23

A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is :

A

{3 \over 4}

B

{3 \over 10}

C

{2 \over 5}

D

{1 \over 5}

Correct Answer

Option C

Solution

If we follow path 1, then probability of getting 1st ball black

= {6 \over {10}}

and probability of getting 2nd ball red when there is 4 R and 8 B balls =

{4 \over {12}}

. So, the probability of getting 1st ball black and 2nd ball red =

{6 \over {10}} \times {4 \over {12}}

. If we follow path 2, then the probability of getting 1st ball red

= {4 \over {10}}

and probability of getting 2nd ball red when in the bag there is 6 red and 6 black balls =

{6 \over {12}}

\therefore\,\,\,

Probability of getting 2nd ball as red

= {6 \over {10}} \times {4 \over {12}} + {4 \over {10}} \times {6 \over {12}}

= {1 \over 5} + {1 \over 5}

= {2 \over 5}

Q24

For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate solve any problem is

{4 \over 5}

, then the probability that he is unable to solve less than two problems is :

A

{{164} \over {25}}{\left( {{1 \over 5}} \right)^{48}}

B

{{316} \over {25}}{\left( {{4 \over 5}} \right)^{48}}

C

{{201} \over 5}{\left( {{1 \over 5}} \right)^{49}}

D

{{54} \over 5}{\left( {{4 \over 5}} \right)^{49}}

Correct Answer

Option D

Solution

There are 50 questions in an exam. So Probability of each question to be correct is p =

{4 \over 5}

and also probability of each question to be incorrect is q =

{1 \over 5}

Let Y is the number of correct question in 50 questions, hence required probability is

{\left( {{4 \over 5}} \right)^{50}} + {}^{50}{C_1}\left( {{1 \over 5}} \right){\left( {{4 \over 5}} \right)^{49}}

=

{\left( {{4 \over 5}} \right)^{49}} + \left( {{4 \over 5} + 10} \right)\left( {{4 \over 5}} \right)

=

{{54} \over 5}{\left( {{4 \over 5}} \right)^{49}}

Q25

A person throws two fair dice. He wins Rs. 15 for throwing a doublet (same numbers on the two dice), wins Rs. 12 when the throw results in the sum of 9, and loses Rs. 6 for any other outcome on the throw. Then the expected gain/loss (in Rs.) of the person is :

A

{1 \over 4}

loss

B

{1 \over 2}

gain

C

{1 \over 2}

loss

D 2 gain

Correct Answer

Option C

Solution

When two dice are thrown then sample space will {(1, 1), (2, 2) ....... (6, 6)} contain total 36 elements number of cases.

Then the expectation will be

{6 \over {36}} \times 15 \times {4 \over {36}} \times 12 - {{26} \over {36}} \times 6

{{90 + 48 - 156} \over {36}} = - {1 \over 2}

=

{1 \over 2}

loss

Q26

If three of the six vertices of a regular hexagon are chosen at random, then the probability that the triangle formed with these chosen vertices is equilateral is :

A

{1 \over {10}}

B

{3 \over {10}}

C

{3 \over {20}}

D

{1 \over {5}}

Correct Answer

Option A

Solution

Choosing vertices of a regular hexagon alternate, here A1, A3, A5 or A2, A4, A6 will result in an equilateral triangle.

Hence the required probability =

{2 \over {{}^6{C_3}}}

=

{1 \over {10}}

Q27

Let a random variable X have a binomial distribution with mean 8 and variance 4. If

P\left( {X \le 2} \right) = {k \over {{2^{16}}}}

, then k is equal to :

A 17

B 1

C 137

D 121

Correct Answer

Option C

Solution

Let number of trials be n and probability of success = p, probability of failure = q Given np = 8, npq = 4 $\Rightarrow$ q =

{1 \over 2}

, p =

{1 \over 2}

, n = 16 (as p + q = 1) p

\left( {x \le 2} \right)

=

{{{}^{16}{C_0} + {}^{16}{C_1} + {}^{16}{C_2}} \over {{2^{16}}}} = {{1 + 16 + 120} \over {{2^{16}}}} = {{137} \over {{2^{16}}}}

Q28

Assume that each born child is equally likely to be a boy or a girl. If two families have two children each, then the conditional probability that all children are girls given that at least two are girls is :

A

{1 \over {10}}

B

{1 \over {17}}

C

{1 \over {11}}

D

{1 \over {12}}

Correct Answer

Option C

Solution

A = At least two girls B = All girls

P\left( {{B \over A}} \right) = {{P\left( {B \cap A} \right)} \over {P\left( A \right)}}

\Rightarrow {{P(B)} \over {P(A)}} = {{{{\left( {{1 \over 4}} \right)}^2}} \over {1 - {}^4{C_0}{{\left( {{1 \over 2}} \right)}^4} - {}^4{C_1}{{\left( {{1 \over 2}} \right)}^4}}}

\Rightarrow {1 \over {16 - 1 - 4}} = {1 \over {11}}

Q29

The minimum number of times one has to toss a fair coin so that the probability of observing at least one head is at least 90% is :

A 2

B 3

C 4

D 5

Correct Answer

Option C

Solution

Probablity of getting head P(H) =

{1 \over 2}

Probablity of getting tail P(T) = 1 -

{1 \over 2}

=

{1 \over 2}

Probability of observing at least one head out of n tosses = 1 - Probability of observing no head occurs out of n tosses = 1 -

{\left( {{1 \over 2}} \right)^n}

According to the question, 1 -

{\left( {{1 \over 2}} \right)^n}

$\ge$

{{90} \over {100}}

$\Rightarrow$

{\left( {{1 \over 2}} \right)^n} \le {1 \over {10}}

$\Rightarrow$

{2^n} \ge 10

As 23 = 8 24 = 16 $\therefore$ Minimum value of n = 4

Q30

Let A, B and C be three events, which are pair-wise independent and

\overrightarrow E

denotes the completement of an event E. If

P\left( {A \cap B \cap C} \right) = 0

and

P\left( C \right) > 0,

then

P\left[ {\left( {\overline A \cap \overline B } \right)\left| C \right.} \right]

is equal to :

A

P\left( {\overline A } \right) - P\left( B \right)

B

P\left( A \right) + P\left( {\overline B } \right)

C

P\left( {\overline A } \right) - P\left( {\overline B } \right)

D

P\left( {\overline A } \right) + P\left( {\overline B } \right)

Correct Answer

Option A

Solution

Here,

P\left( {\overline A \cap \overline B \left| C \right.} \right) = {{P\left( {\overline A \cap \overline B \cap C} \right)} \over {P\left( C \right)}}

=

{{P\left[ {\left( {\overline {A \cup B} } \right) \cap C} \right]} \over {P\left( C \right)}}

=

{{P\left[ {C - \left( {A \cup B} \right)} \right]} \over {P\left( C \right)}}

=

{{P\left( C \right) - P\left( {A \cap C} \right) - P\left( {B \cap C} \right) + P\left( {A \cap B \cap C} \right)} \over {P\left( C \right)}}

=

{{P\left( C \right) - P\left( {A \cap C} \right) - P\left( {B \cap C} \right)} \over {P\left( C \right)}}

( $\because$

\left. {P\left( {A \cap B \cap C} \right) = 0} \right)

=

{{P\left( C \right) - P\left( A \right).P(C) - P\left( B \right).P(C)} \over {P\left( C \right)}}

[ $\because$ A, B and C are independent events] = 1 - P(A) - P(B) =

P\left( {\overline A } \right)

- P(B) or

P\left( {\overline B } \right)

- P(A)