Probability — Page 8 | JEE Mathematics

Q71

Let A and B be independent events such that P(A) = p, P(B) = 2p. The largest value of p, for which P (exactly one of A, B occurs) =

{5 \over 9}

, is :

A

{1 \over 3}

B

{2 \over 9}

C

{4 \over 9}

D

{5 \over 12}

Correct Answer

Option D

Solution

P (Exactly one of A or B)

= P\left( {A \cap \overline B } \right) + \left( {\overline A \cap B} \right) = {5 \over 9}

= P(A)P(\overline B ) + P(\overline A )P(B) = {5 \over 9}

\Rightarrow P(A)(1 - P(B)) + (1 - P(A))P(B) = {5 \over 9}

\Rightarrow p(1 - 2p) + (1 - p)2p = {5 \over 9}

\Rightarrow 36{p^2} - 27p + 5 = 0

\Rightarrow p = {1 \over 3}

or

{5 \over {12}}

{p_{\max }} = {5 \over {12}}

Q72

A fair die is tossed until six is obtained on it. Let x be the number of required tosses, then the conditional probability P(x

\ge

5 | x > 2) is :

A

{{125} \over {216}}

B

{{11} \over {36}}

C

{{5} \over {6}}

D

{{25} \over {36}}

Correct Answer

Option D

Solution

P(x $\ge$ 5 | x > 2) =

{{P(x \ge 5)} \over {P(x > 2)}}

=

{{{{\left( {{5 \over 6}} \right)}^4}.{1 \over 6} + {{\left( {{5 \over 6}} \right)}^5}.{1 \over 6} + ....... + \infty } \over {{{\left( {{5 \over 6}} \right)}^2}.{1 \over 6} + {{\left( {{5 \over 6}} \right)}^3}.{1 \over 6} + ...... + \infty }}

=

{{{{{{\left( {{5 \over 6}} \right)}^4}.{1 \over 5}} \over {1 - {5 \over 6}}}} \over {{{{{\left( {{5 \over 6}} \right)}^2}.{1 \over 6}} \over {1 - {5 \over 6}}}}} = {\left( {{5 \over 6}} \right)^2} = {{25} \over {36}}

Q73

Two fair dice are thrown. The numbers on them are taken as

\lambda

and

\mu

, and a system of linear equations x + y + z = 5 x + 2y + 3z =

\mu

x + 3y +

\lambda

z = 1 is constructed. If p is the probability that the system has a unique solution and q is the probability that the system has no solution, then :

A

p = {1 \over 6}

and

q = {1 \over 36}

B

p = {5 \over 6}

and

q = {5 \over 36}

C

p = {5 \over 6}

and

q = {1 \over 36}

D

p = {1 \over 6}

and

q = {5 \over 36}

Correct Answer

Option B

Solution

D \ne 0 \Rightarrow \left| \begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda \end{array} \right| \ne 0 \Rightarrow \lambda \ne 5

For no solution D = 0 $\Rightarrow$ $\lambda$ = 5

{D_1} = \left| \begin{array}{lll}1 & 1 & 5 \\ 1 & 2 & \mu \\ 1 & 3 & 1 \end{array} \right| \ne 0 \Rightarrow \mu \ne 3

p = {5 \over 6}

q = {1 \over 6} \times {5 \over 6} = {5 \over {36}}

Option (b).

Q74

Bag A contains 3 white, 7 red balls and Bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag A, if the ball drawn is white, is

A 1/4

B 1/3

C 3/10

D 1/9

Correct Answer

Option B

Solution

\mathrm{E}_1: \mathrm{A}

is selected

\mathrm{E}_2: \mathrm{B}

is selected

\mathrm{E}

: white ball is drawn

\mathrm{P}\left(\mathrm{E}_1 / \mathrm{E}\right)=

\begin{aligned} & \frac{P(E) \cdot P\left(E / E_1\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right)}=\frac{\frac{1}{2} \times \frac{3}{10}}{\frac{1}{2} \times \frac{3}{10}+\frac{1}{2} \times \frac{3}{5}} \\ & =\frac{3}{3+6}=\frac{1}{3} \end{aligned}

Q75

When a certain biased die is rolled, a particular face occurs with probability

{1 \over 6} - x

and its opposite face occurs with probability

{1 \over 6} + x

. All other faces occur with probability

{1 \over 6}

. Note that opposite faces sum to 7 in any die. If 0 < x <

{1 \over 6}

, and the probability of obtaining total sum = 7, when such a die is rolled twice, is

{13 \over 96}

, then the value of x is :

A

{1 \over 16}

B

{1 \over 8}

C

{1 \over 9}

D

{1 \over 12}

Correct Answer

Option B

Solution

Probability of obtaining total sum 7 = probability of getting opposite faces. Probability of getting opposite faces

= 2\left[ {\left( {{1 \over 6} - x} \right)\left( {{1 \over 6} + x} \right) + {1 \over 6} \times {1 \over 6} + {1 \over 6} \times {1 \over 6}} \right]

\Rightarrow 2\left[ {\left( {{1 \over 6} - x} \right)\left( {{1 \over 6} + x} \right) + {1 \over 6} \times {1 \over 6} + {1 \over 6} \times {1 \over 6}} \right] = {{13} \over {96}}

(given) $\Rightarrow$

x = {1 \over 8}

Q76

Each of the persons A and B independently tosses three fair coins. The probability that both of them get the same number of heads is :

A

{1 \over 8}

B

{5 \over 8}

C

{5 \over 16}

D 1

Correct Answer

Option C

Solution

Let x be the number of heads obtained by A, and y be the number of heads obtained by B.

Note that x and y are binomial variable with parameters n = 3 and p =

{1 \over 2}

$\therefore$ Probability that both A and B obtained the same number of heads is

= P(x = 0)\,.\,P(y = 0) + P(x = 1)\,.\,P(y = 1) + P(x = 2)\,.\,P(y = 2) + P(x = 3)\,.\,P(y = 3)

= {\left[ {{3_{{C_0}}}{{\left( {{1 \over 2}} \right)}^3}} \right]^2} + {\left[ {{3_{{C_1}}}{{\left( {{1 \over 2}} \right)}^3}} \right]^2} + {\left[ {{3_{{C_2}}}{{\left( {{1 \over 2}} \right)}^3}} \right]^2} + {\left[ {{3_{{C_3}}}{{\left( {{1 \over 2}} \right)}^3}} \right]^2}

= {\left( {{1 \over 2}} \right)^6}\left[ {1 + 9 + 9 + 1} \right]

= {{20} \over {64}}

= {5 \over {16}}

Q77

Let S = {1, 2, 3, 4, 5, 6}. Then the probability that a randomly chosen onto function g from S to S satisfies g(3) = 2g(1) is :

A

{1 \over {10}}

B

{1 \over {15}}

C

{1 \over {5}}

D

{1 \over {30}}

Correct Answer

Option A

Solution

g(3) = 2g(1) can be defined in 3 ways number of onto functions in this condition = 3 $\times$ 4!

Total number of onto functions = 6!

Required probability =

{{3 \times 4!} \over {6!}} = {1 \over {10}}

Q78

The probability that a randomly chosen 2

\times

2 matrix with all the entries from the set of first 10 primes, is singular, is equal to :

A

{{133} \over {{{10}^4}}}

B

{{18} \over {{{10}^3}}}

C

{{19} \over {{{10}^3}}}

D

{{271} \over {{{10}^4}}}

Correct Answer

Option C

Solution

First 10 prime numbers are ={2, 3, 5, 7, 11, 13, 17, 19, 23, 29} Let A is a 2 $\times$ 2 matrix,

A = \left[ \begin{array}{ll}a & b \\ c & d \end{array} \right]

Given that matrix A is singular. $\therefore$ | A | = 0

\Rightarrow \left| \begin{array}{ll}a & b \\ c & d \end{array} \right| = 0

\Rightarrow ad = bc

Case I : ad = bc condition satisfy when a = b = c = d.

For ex when a = 2, b = 2, c = 2, d = 2, then ad = bc satisfy.

Now there are 10 prime numbers.

We can choose any one of the 10 prime number in

{}^{10}C_{1}

= 10 ways and put them in the four positions of the matrix and matrix will be singular.

$\therefore$ In this case, total favorable case = 10 Case 2 : ad = bc condition satisfies when (1) a = 2, d - 3 then (a) b = 2, c = 3 (b) b = 3, c = 2 or a = 3, d = 2 then (a) b = 2, c = 3 (b) b = 3, c = 2 So you can see for two different prime number for a and d there are 4 possible value of b and c which satisfy ad = bc condition.

Two different values of a and d can be chosen from 10 prime numbers =

{}^{10}C_{2}

ways And for each combination of a and d there are 4 possible values of b and c. $\therefore$ Total possible values =

{}^{10}C_{2}

$\times$ 4 From case I and case II total possible values of 10 prime numbers which satisfy ad = bc condition = 10 +

{}^{10}C_{2}

$\times$ 4 For sample space, Number of ways to fill element a of matrix A = chose any prime number among 10 available prime number =

{}^{10}C_{1}

ways Similarly, For element b of matrix A =

{}^{10}C_{1}

ways For element c of matrix A =

{}^{10}C_{1}

ways For element d of matrix A =

{}^{10}C_{1}

ways $\therefore$ Sample space =

{}^{10}C_{1}

$\times$

{}^{10}C_{1}

$\times$

{}^{10}C_{1}

$\times$

{}^{10}C_{1}

= 104 $\therefore$ Probability

= {{10 + {{}^{10}C_{2}} \times 4} \over {{{10}^4}}}

= {{10 + 180} \over {{{10}^4}}}

= {{190} \over {{{10}^4}}}

= {{19} \over {{{10}^3}}}

Q79

The probability that a relation R from {x, y} to {x, y} is both symmetric and transitive, is equal to :

A

{5 \over {16}}

B

{9 \over {16}}

C

{11 \over {16}}

D

{13 \over {16}}

Correct Answer

Option A

Solution

Total number of relations $=2^{2^{2}}=2^{4}=16$ Relations that are symmetric as well as transitive are $\phi,\{(x, x)\},\{(y, y)\},\{(x, x),(x, y),(y, y),(y, x)\},\{(x, x),(y, y)\}$ $\therefore \quad$ favourable cases $=5$ $\therefore \quad P_{r}=\dfrac{5}{16}$

Q80

The probability that a randomly chosen one-one function from the set {a, b, c, d} to the set {1, 2, 3, 4, 5} satisfies f(a) + 2f(b)

-

f(c) = f(d) is :

A

{1 \over {24}}

B

{1 \over {40}}

C

{1 \over {30}}

D

{1 \over {20}}

Correct Answer

Option D

Solution

Number of one-one function from $\{a, b, c, d\}$ to set $\{1,2,3,4,5\}$ is ${ }^{5} P_{4}=120 n(s)$ .

The required possible set of value (f(a), $f(b), f(c), f(d))$ such that $f(a)+2 f(b)-f(c)=f(d)$ are $(5,3,2,1),(5,1,2,3),(4,1,3,5),(3,1,4,5)$ , $(5,4,3,2)$ and $(3,4,5,2)$ $\therefore n(E)=6$ $\therefore$ Required probability $=\dfrac{n(E)}{n(S)}=\dfrac{6}{120}=\dfrac{1}{20}$