Probability — Page 9 | JEE Mathematics

Q81

The probability, that in a randomly selected 3-digit number at least two digits are odd, is :

A

{{19} \over {36}}

B

{{15} \over {36}}

C

{{13} \over {36}}

D

{{23} \over {36}}

Correct Answer

Option A

Solution

At least two digits are odd = exactly two digits are odd + exactly there 3 digits are odd For exactly three digits are odd For exactly two digits odd : If 0 is used then $: 2 \times 5 \times 5=50$ If 0 is not used then : ${ }^3 \mathrm{C}_1 \times 4 \times 5 \times 5=300$ Required Probability $=\dfrac{475}{900}=\dfrac{19}{36}$

Q82

If a point A(x, y) lies in the region bounded by the y-axis, straight lines 2y + x = 6 and 5x

-

6y = 30, then the probability that y

A

{1 \over 6}

B

{5 \over 6}

C

{2 \over 3}

D

{6 \over 7}

Correct Answer

Option B

Solution

The required probability

= {\text{Area of Region PQCAP} \over \text{Area of Region ABCA}}

= {{{1 \over 2} \times 8 \times 6 - {1 \over 2} \times 2 \times 4} \over {{1 \over 2} \times 8 \times 6}}

= {5 \over 6}

Q83

Five numbers

{x_1},{x_2},{x_3},{x_4},{x_5}

are randomly selected from the numbers 1, 2, 3, ......., 18 and are arranged in the increasing order $$({x_1}

A

{1 \over {136}}

B

{1 \over {72}}

C

{1 \over {68}}

D

{1 \over {34}}

Correct Answer

Option C

Solution

No. of ways to select and arrange $\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4, \mathrm{x}_5$ from $1,2,3$ .......18

\begin{aligned} & \mathrm{n}(\mathrm{s})={ }^{18} \mathrm{C}_5 \\\\ & \begin{array}{lllll} x_1 & \underset{7}{\left(x_2\right)} & x_3 & \underset{11}{\left(x_4\right)} & x_5 \end{array} \\\\ & \mathrm{n}(\mathrm{E})={ }^6 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1 \times{ }^7 \mathrm{C}_1 \\\\ & P(E)=\frac{6 \times 3 \times 7}{{ }^{18} C_5} \\\\ & \frac{1}{17 \times 4}=\frac{1}{68} \\\\ \end{aligned}

Q84

Two integers

x

and

y

are chosen with replacement from the set

\{0,1,2,3, \ldots, 10\}

. Then the probability that

|x-y|>5

, is :

A

\dfrac{31}{121}

B

\dfrac{60}{121}

C

\dfrac{62}{121}

D

\dfrac{30}{121}

Correct Answer

Option D

Solution

If

x=0, y=6,7,8,9,10

If

x=1, y=7,8,9,10

If

x=2, y=8,9,10

If

x=3, y=9,10

If

x=4, y=10

If

x=5, y=

no possible value Total possible ways

=(5+4+3+2+1) \times 2

=30

Required probability

=\frac{30}{11 \times 11}=\frac{30}{121}

Q85

A six faced die is biased such that

3 \times \mathrm{P}(

a prime number

)\,=6 \times \mathrm{P}(

a composite number

)\,=2 \times \mathrm{P}(1)

. Let X be a random variable that counts the number of times one gets a perfect square on some throws of this die. If the die is thrown twice, then the mean of X is :

A

\dfrac{3}{11}

B

\dfrac{5}{11}

C

\dfrac{7}{11}

D

\dfrac{8}{11}

Correct Answer

Option D

Solution

Let P(a prime number) = $\alpha$ P(a composite number) = $\beta$ and P(1) = $\gamma$ $\because$

3\alpha = 6\beta = 2\gamma = k

(say) and

3\alpha + 2\beta + \gamma = 1

\Rightarrow k + {k \over 3} + {k \over 2} = 1 \Rightarrow k = {6 \over {11}}

Mean = np where n = 2 and p = probability of getting perfect square

= P(1) + P(4) = {k \over 2} + {k \over 6} = {4 \over {11}}

So, mean

= 2\,.\,\left( {{4 \over {11}}} \right) = {8 \over {11}}

Q86

Let a biased coin be tossed 5 times. If the probability of getting 4 heads is equal to the probability of getting 5 heads, then the probability of getting atmost two heads is :

A

{{275} \over {{6^5}}}

B

{{36} \over {{5^4}}}

C

{{181} \over {{5^5}}}

D

{{46} \over {{6^4}}}

Correct Answer

Option D

Solution

Coin is tossed 5 times, so n = 5 Let, p = probability of getting heads q = probability of getting tails. $\therefore$ p + q = 1 ......

(1) $\therefore$ Probability of getting 4 heads = 5C4 . p4 . q And probability of getting 5 heads = 5C5 . p5 Given, 5C4 . p4 . q = 5C5 . p5 $\Rightarrow$ 5q = p .......

(2) From equation (1) and (2), we get, 5q + q = 1 $\Rightarrow$ 6q = 1 $\Rightarrow$ q =

{1 \over 6}

$\therefore$ p = 1 $-$

{1 \over 6}

=

{5 \over 6}

Now, probability of getting atmost two heads = p (x = 0) + p (x = 1) + p (x = 2) p (x = 0) = Getting zero head in 5 trials = 5C0 . p0 . q5 p (x = 1) = Getting one head in 5 trials = 5C1 . p1 . q4 p (x = 2) = Getting two heads in 5 trials = 5C2 . p2 . q3 = 5C0 . q5 + 5C1 . pq4 + 5C2 . p2q3

= {\left( {{1 \over 6}} \right)^5} + 5\,.\,{5 \over 6}\,.\,{\left( {{1 \over 6}} \right)^4} + 10\,.\,{\left( {{5 \over 6}} \right)^2}\,.\,{\left( {{1 \over 6}} \right)^3}

= {{1 + 25 + 250} \over {{6^5}}} = {{276} \over {{6^5}}}

=

{{46} \over {{6^4}}}

Q87

A biased die is marked with numbers 2, 4, 8, 16, 32, 32 on its faces and the probability of getting a face with mark n is

{1 \over n}

. If the die is thrown thrice, then the probability, that the sum of the numbers obtained is 48, is :

A

{7 \over {{2^{11}}}}

B

{7 \over {{2^{12}}}}

C

{3 \over {{2^{10}}}}

D

{{13} \over {{2^{12}}}}

Correct Answer

Option D

Solution

There are only two ways to get sum 48, which are (32, 8, 8) and (16, 16, 16) So, required probability

= 3\left( {{2 \over {32}}\,.\,{1 \over 8}\,.\,{1 \over 8}} \right) + \left( {{1 \over {16}}\,.\,{1 \over {16}}\,.\,{1 \over {16}}} \right)

= {3 \over {{2^{10}}}} + {1 \over {{2^{12}}}}

= {{13} \over {{2^{12}}}}

Q88

Let E1 and E2 be two events such that the conditional probabilities

P({E_1}|{E_2}) = {1 \over 2}

,

P({E_2}|{E_1}) = {3 \over 4}

and

P({E_1} \cap {E_2}) = {1 \over 8}

. Then :

A

P({E_1} \cap {E_2}) = P({E_1})\,.\,P({E_2})

B

P(E{'_1} \cap E{'_2}) = P(E{'_1})\,.\,P(E{_2})

C

P({E_1} \cap E{'_2}) = P({E_1})\,.\,P({E_2})

D

P(E{'_1} \cap {E_2}) = P({E_1})\,.\,P({E_2})

Correct Answer

Option C

Solution

P\left( {{{{E_1}} \over {{E_2}}}} \right) = {1 \over 2} \Rightarrow {{P({E_1} \cap {E_2})} \over {P({E_2})}} = {1 \over 2}

P\left( {{{{E_2}} \over {{E_1}}}} \right) = {3 \over 4} \Rightarrow {{P({E_2} \cap {E_1})} \over {P({E_1})}} = {3 \over 4}

P({E_1} \cap {E_2}) = {1 \over 8}

P({E_2}) = {1 \over 4},\,P({E_1}) = {1 \over 6}

(A)

P({E_1} \cap {E_2}) = {1 \over 8}

and

P({E_1})\,.\,P({E_2}) = {1 \over {24}}

\Rightarrow P({E_1} \cap {E_2}) \ne P({E_1})\,.\,P({E_2})

(B)

P(E{'_1} \cap E{'_2}) = 1 - P({E_1} \cup {E_2})

= 1 - \left[ {{1 \over 4} + {1 \over 6} - {1 \over 8}} \right] = {{17} \over {24}}

P(E{'_1}) = {3 \over 4} \Rightarrow P(E{'_1})P({E_2}) = {3 \over {24}}

\Rightarrow P(E{'_1} \cap E{'_2}) \ne P(E{'_1})\,.\,P({E_2})

(C)

P({E_1} \cap E{'_2}) = P({E_1}) - P({E_1} \cap {E_2})

= {1 \over 6} - {1 \over 8} = {1 \over {24}}

P({E_1})\,.\,P({E_2}) = {1 \over {24}}

\Rightarrow P({E_1} \cap E{'_2}) = P({E_1})\,.\,P({E_2})

(D)

P(E{'_1} \cap {E_2}) = P({E_2}) - P({E_1} \cap {E_2})

= {1 \over 4} - {1 \over 8} = {1 \over 8}

P({E_1})P({E_2}) = {1 \over {24}}

\Rightarrow P(E{'_1} \cap {E_2}) \ne P({E_1})\,.\,P({E_2})

Q89

Bag A contains 2 white, 1 black and 3 red balls and bag B contains 3 black, 2 red and n white balls. One bag is chosen at random and 2 balls drawn from it at random, are found to be 1 red and 1 black. If the probability that both balls come from Bag A is

{6 \over {11}}

, then n is equal to __________.

A 13

B 6

C 4

D 3

Correct Answer

Option C

Solution

\begin{aligned} &P(1 R \text{ and } 1 B)=P(A) \cdot P\left(\frac{1 R 1 B}{A}\right)+P(B) \cdot P\left(\frac{1 R 1 B}{B}\right) \\\\ &=\frac{1}{2} \cdot \frac{{ }^{3} C_{1} \cdot{ }^{1} C_{1}}{{ }^{6} C_{2}}+\frac{1}{2} \cdot \frac{{ }^{2} C_{1} \cdot{ }^{3} C_{1}}{{ }^{n+5} C_{2}} \\\\ &P\left(\frac{1 R 1 B}{A}\right)=\frac{\frac{1}{2} \cdot \frac{3}{15}}{\frac{1}{2} \cdot \frac{3}{15}+\frac{1}{2} \cdot \frac{6 \cdot 2}{(n+5)(n+4)}}=\frac{6}{11} \\\\ &\Rightarrow \frac{\frac{1}{10}}{\frac{1}{10}+\frac{6}{(n+5)(n+4)}}=\frac{6}{11} \end{aligned}

\begin{aligned} &\Rightarrow \frac{11}{10}=\frac{6}{10}+\frac{36}{(n+5)(n+4)} \\\\ &\Rightarrow \frac{5}{10 \times 36}=\frac{1}{(n+5)(n+4)} \\\\ &\Rightarrow n^{2}+9 n-52=0 \\\\ &\Rightarrow n=4 \text{ is only possible value } \end{aligned}

Q90

If a random variable X follows the Binomial distribution B(33, p) such that

3P(X = 0) = P(X = 1)

, then the value of

{{P(X = 15)} \over {P(X = 18)}} - {{P(X = 16)} \over {P(X = 17)}}

is equal to :

A 1320

B 1088

C

{{120} \over {1331}}

D

{{1088} \over {1089}}

Correct Answer

Option A

Solution

$3 P(X=0)=P(X=1)$

\begin{aligned} &3 \cdot{ }^{n} C_{0} P^{0}(1-P)^{n}={ }^{n} C_{1} P^{1}(1-P)^{n-1} \\\\ &\frac{3}{n}=\frac{P}{1-P} \Rightarrow \frac{1}{11}=\frac{P}{1-P} \\\\ &\Rightarrow 1-P=11 P \\\\ &\Rightarrow P=\frac{1}{12} \end{aligned}

\frac{P(X=15)}{P(X=18)}-\frac{P(X=16)}{P(X=17)}

\begin{aligned} &\Rightarrow \frac{{ }^{33} C_{15} P^{15}(1-P)^{18}}{{ }^{33} C_{18} P^{18}(1-P)^{15}}-\frac{{ }^{33} C_{16} P^{16}(1-P)^{17}}{{ }^{33} C_{17} P^{17}(1-P)^{16}} \\\\ &\Rightarrow\left(\frac{1-P}{P}\right)^{3}-\left(\frac{1-P}{P}\right) \\\\ &\Rightarrow \quad 11^{3}-11=1320 \end{aligned}