Probability — Page 7 | JEE Mathematics

Q61

Let in a Binomial distribution, consisting of 5 independent trials, probabilities of exactly 1 and 2 successes be 0.4096 and 0.2048 respectively. Then the probability of getting exactly 3 successes is equal to :

A

{{40} \over {243}}

B

{{128} \over {625}}

C

{{80} \over {243}}

D

{{32} \over {625}}

Correct Answer

Option D

Solution

{}^5{C_1}{p^1}{q^4}

= 0.4096 ..... (1)

{}^5{C_2}{p^2}{q^3}

= 0.2048 ..... (2)

{{(1)} \over {(2)}} \Rightarrow {q \over {2p}} = 2 \Rightarrow q = 4p

p + q = 1 \Rightarrow P = {1 \over 5},q = {4 \over 5}

P (exactly 3) =

{}^5{C_3}{(p)^3}{(q)^2} = {}^5{C_3}{\left( {{1 \over 5}} \right)^3}{\left( {{4 \over 5}} \right)^2}

= 10 \times {1 \over {125}} \times {{16} \over {25}} = {{32} \over {625}}

Q62

Words with or without meaning are to be formed using all the letters of the word EXAMINATION. The probability that the letter M appears at the fourth position in any such word is :

A

{1 \over {66}}

B

{1 \over {11}}

C

{1 \over {9}}

D

{2 \over {11}}

Correct Answer

Option B

Solution

AAEIIMNNOTX ----------------M---------------- Total words with M at fourth Place =

{{10!} \over {2!2!2!}}

Total words =

{{11!} \over {2!2!2!}}

Required probability =

{{10!} \over {11!}}

=

{{1} \over {11}}

Q63

The probability of selecting integers a

\in

[

-

5, 30] such that x2 + 2(a + 4)x

-

5a + 64 > 0, for all x

\in

R, is :

A

{7 \over {36}}

B

{2 \over {9}}

C

{1 \over {6}}

D

{1 \over {4}}

Correct Answer

Option B

Solution

D < 0 $\Rightarrow$ 4(a + 4)2 $-$ 4( $-$ 5a + 64) < 0 $\Rightarrow$ a2 + 16 + 8a + 5a $-$ 64 < 0 $\Rightarrow$ a2 + 13a $-$ 48 < 0 $\Rightarrow$ (a + 16) (a $-$ 3) < 0 $\Rightarrow$ a

\in

( $-$ 16, 3) $\therefore$ Possible a : { $-$ 5, $-$ 4, ............., 3} $\therefore$ Required probability =

{8 \over {36}}

=

{2 \over {9}}

Q64

Let A, B and C be three events such that the probability that exactly one of A and B occurs is (1

-

k), the probability that exactly one of B and C occurs is (1

-

2k), the probability that exactly one of C and A occurs is (1

-

k) and the probability of all A, B and C occur simultaneously is k2, where 0 < k < 1. Then the probability that at least one of A, B and C occur is :

A greater than

{1 \over 8}

but less than

{1 \over 4}

B greater than

{1 \over 2}

C greater than

{1 \over 4}

but less than

{1 \over 2}

D exactly equal to

{1 \over 2}

Correct Answer

Option B

Solution

P(\overline A \cap B) + P(A \cap \overline B ) = 1 - k

P(\overline A \cap C) + P(A \cap \overline C ) = 1 - 2k

P(\overline B \cap C) + P(B \cap \overline C ) = 1 - k

P(A \cap B \cap C) = {k^2}

P(A) + P(B) - 2P(A \cap B) = 1 - k

.....(i)

P(B) + P(C) - 2P(B \cap C) = 1 - k

..... (ii)

P(C) + P(A) - 2P(A \cap C) = 1 - 2k

..... (iii)

(i) + (ii) + (iii)

P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) = {{ - 4k + 3} \over 2}

So,

P(A \cup B \cup C) = {{ - 4k + 3} \over 2} + {k^2}

P(A \cup B \cup C) = {{2{k^2} - 4k + 3} \over 2}

= {{2{{(k - 1)}^2} + 1} \over 2}

P(A \cup B \cup C) > {1 \over 2}

Q65

Let X be a random variable having binomial distribution B(7, p). If P(X = 3) = 5P(x = 4), then the sum of the mean and the variance of X is :

A

{105 \over {16}}

B

{7\over {16}}

C

{77\over {36}}

D

{49\over {16}}

Correct Answer

Option C

Solution

Given P(X = 3) = 5P(X = 4) and n = 7

\Rightarrow {}^7{C_3}{p^3}{q^4} = 5\,.\, \Rightarrow {}^7{C_4}{p^4}{q^3}

\Rightarrow q = 5p

and also

p + q = 1

\Rightarrow p = {1 \over 6}

and

q = {5 \over 6}

Mean

= {7 \over 6}

and variance

= {{35} \over {36}}

Mean + Variance

= {7 \over 6} + {{35} \over {36}} = {{77} \over {36}}

Q66

Four dice are thrown simultaneously and the numbers shown on these dice are recorded in 2

\times

2 matrices. The probability that such formed matrix have all different entries and are non-singular, is :

A

{{45} \over {162}}

B

{{21} \over {81}}

C

{{22} \over {81}}

D

{{43} \over {162}}

Correct Answer

Option D

Solution

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

| A | = ad $-$ bc Total case = 64 For non-singular matrix | A | $\ne$ 0 $\Rightarrow$ ad $-$ bc $\ne$ 0 $\Rightarrow$ ad $\ne$ bc And a, b, c, d are all different numbers in the set {1, 2, 3, 4, 5, 6} Now for ad = bc (i) 6 $\times$ 1 = 2 $\times$ 3 $\Rightarrow$

\left. \begin{array}{ll}a = 6,b = 2,c = 3,d = 1 \\ or\,a = 1,b = 2,c = 3,d = 6 \\ : \\ : \end{array} \right\}

8 each cases (ii) 6 $\times$ 2 = 3 $\times$ 4 $\Rightarrow$

\left. \begin{array}{ll}a = 6,b = 3,c = 4,d = 2 \\ or\,a = 2,b = 3,c = 4,d = 6 \\ : \\ : \end{array} \right\}

8 such cases favourable cases =

^6C_4 \times 4! - 16

required probability

= {{{^6C_4 \times 4!} - 16} \over {{6^4}}} = {{43} \over {162}}

Q67

Let 9 distinct balls be distributed among 4 boxes, B1, B2, B3 and B4. If the probability than B3 contains exactly 3 balls is

k{\left( {{3 \over 4}} \right)^9}

then k lies in the set :

A {x

\in

R : |x

-

3| < 1}

B {x

\in

R : |x

-

2|

\le

1}

C {x

\in

R : |x

-

1| < 1}

D {x

\in

R : |x

-

5|

\le

1}

Correct Answer

Option A

Solution

Required probability =

{{{}^9{C_3}{{.3}^6}} \over {{4^9}}}

= {{{}^9{C_3}} \over {27}}.{\left( {{3 \over 4}} \right)^9}

= {{28} \over 9}.{\left( {{3 \over 4}} \right)^9} \Rightarrow k = {{28} \over 9}

Which satisfies

\left| {x - 3} \right| < 1

Q68

A student appeared in an examination consisting of 8 true-false type questions. The student guesses the answers with equal probability. the smallest value of n, so that the probability of guessing at least 'n' correct answers is less than

{1 \over 2}

, is :

A 5

B 6

C 3

D 4

Correct Answer

Option A

Solution

P(E) < {1 \over 2}

\Rightarrow \sum\limits_{r = n}^8 {{}^8{C_r}} {\left( {{1 \over 2}} \right)^{8 - r}}{\left( {{1 \over 2}} \right)^r} < {1 \over 2}

\Rightarrow \sum\limits_{r = n}^8 {{}^8{C_r}} {\left( {{1 \over 2}} \right)^8} < {1 \over 2}

\Rightarrow {}^8{C_n} + {}^8{C_{n + 1}} + .... + {}^8{C_8} < 128

\Rightarrow 256 - \left( {{}^8{C_0} + {}^8{C_1} + ... + {}^8{C_{n - 1}}} \right) < 128

\Rightarrow {}^8{C_0} + {}^8{C_1} + .... + {}^8{C_{n - 1}} > 128

\Rightarrow n - 1 \ge 4

\Rightarrow n \ge 5

Q69

Let X be a random variable such that the probability function of a distribution is given by

P(X = 0) = {1 \over 2},P(X = j) = {1 \over {{3^j}}}(j = 1,2,3,...,\infty )

. Then the mean of the distribution and P(X is positive and even) respectively are :

A

{3 \over 8}

and

{1 \over 8}

B

{3 \over 4}

and

{1 \over 8}

C

{3 \over 4}

and

{1 \over 9}

D

{3 \over 4}

and

{1 \over 16}

Correct Answer

Option B

Solution

Mean =

\sum {{X_i}{P_i}} = \sum\limits_{r = 0}^\infty {r.{1 \over {{3^r}}} = {3 \over 4}}

P(X is even)

= {1 \over {{3^2}}} + {1 \over {{3^4}}} + ...\infty

= {{{1 \over 9}} \over {1 - {1 \over 9}}} = {{1/9} \over {8/9}} = {1 \over 8}

Q70

The probability that a randomly selected 2-digit number belongs to the set {n

\in

N : (2n

-

2) is a multiple of 3} is equal to :

A

{1 \over 6}

B

{2 \over 3}

C

{1 \over 2}

D

{1 \over 3}

Correct Answer

Option C

Solution

Total number of cases =

{}^{90}{C_1} = 90

Now,

{2^n} - 2 = {(3 - 1)^n} - 2

{}^n{C_0}{3^n} - {}^n{C_1}{.3^{n - 1}} + .... + {( - 1)^{n - 1}}.{}^n{C_{n - 1}}3 + {( - 1)^n}.{}^n{C_n} - 2

=

3\left( {{3^{n - 1}} - n{3^{n - 2}} + ... + {{( - 1)}^{n - 1}}.n} \right) + {( - 1)^n} - 2

({2^n} - 2)

is multiply of 3 only when n is odd Required Probability

= {{45} \over {90}} = {1 \over 2}