Probability — Page 6 | JEE Mathematics

Q51

An ordinary dice is rolled for a certain number of times. If the probability of getting an odd number 2 times is equal to the probability of getting an even number 3 times, then the probability of getting an odd number for odd number of times is :

A

{5 \over {36}}

B

{3 \over {16}}

C

{1 \over 2}

D

{1 \over {32}}

Correct Answer

Option C

Solution

P(odd no. twice) = P(even no. thrice)

\Rightarrow {}^n{C_2}{\left( {{1 \over 2}} \right)^n} = {}^n{C_3}{\left( {{1 \over 2}} \right)^n} \Rightarrow n = 5

Success is getting an odd number then P(odd successes) = P(1) + P(3) + P(5)

= {}^5{C_1}{\left( {{1 \over 2}} \right)^5} + {}^5{C_3}{\left( {{1 \over 2}} \right)^5} + {}^5{C_5}{\left( {{1 \over 2}} \right)^5}

= {{16} \over {{2^5}}} = {1 \over 2}

Q52

The probability that two randomly selected subsets of the set {1, 2, 3, 4, 5} have exactly two elements in their intersection, is :

A

{{135} \over {{2^9}}}

B

{{65} \over {{2^8}}}

C

{{65} \over {{2^7}}}

D

{{35} \over {{2^7}}}

Correct Answer

Option A

Solution

Given, set P = {1, 2, 3, 4, 5} Let the two subsets be A and B Then, n (A $\cap$ B) = 2 (as given in question) We can choose two elements from set P in 5C2 ways.

After choosing two common elements for set A and B, each of remaining three elements from set P have three choice (1) It can go to set A (2) It can go to set B (3) It don't go to any sets it stays at set P.

$\therefore$ Total ways for the three elements = 3 $\times$ 3 $\times$ 3 = 33 $\therefore$ Required probability =

{{{}^5{C_2} \times {3^3}} \over {\left( {{2^5}} \right)\left( {{2^5}} \right)}}

=

{{{}^5{C_2} \times {3^3}} \over {{4^5}}} = {{10 \times 27} \over {{2^{10}}}} = {{135} \over {{2^9}}}

Q53

When a missile is fired from a ship, the probability that it is intercepted is

{1 \over 3}

and the probability that the missile hits the target, given that it is not intercepted, is

{3 \over 4}

. If three missiles are fired independently from the ship, then the probability that all three hit the target, is :

A

{3 \over 4}

B

{3 \over 8}

C

{1 \over 27}

D

{1 \over 8}

Correct Answer

Option D

Solution

Probability of not getting intercepted =

{2 \over 3}

When it is not intercepted, probability of missile hitting target =

{3 \over 4}

$\therefore$ So when such 3 missiles launched then P (all 3 hitting the target) =

{\left( {{2 \over 3} \times {3 \over 4}} \right)}

$\times$

{\left( {{2 \over 3} \times {3 \over 4}} \right)}

$\times$

{\left( {{2 \over 3} \times {3 \over 4}} \right)}

= {1 \over 8}

Q54

The coefficients a, b and c of the quadratic equation, ax2 + bx + c = 0 are obtained by throwing a dice three times. The probability that this equation has equal roots is :

A

{1 \over {72}}

B

{5 \over {216}}

C

{1 \over {36}}

D

{1 \over {54}}

Correct Answer

Option B

Solution

ax2 + bx + c = 0 a, b, c

\in

{1,2,3,4,5,6} n(s) = 6 × 6 × 6 = 216 For equal roots, D = 0 $\Rightarrow$ b2 = 4ac $\Rightarrow$ ac =

{{{b^2}} \over 4}

Favourable case : If b = 2, ac = 1 $\Rightarrow$ a = 1, c = 1 If b = 4, ac = 4 : a = 1, c = 4 a = 4, c = 1 a = 2, c = 2 If b = 6, ac = 9 $\Rightarrow$ a = 3, c = 3 $\therefore$ Favorable cases = 5 $\therefore$ Required probability =

{5 \over {216}}

Q55

Let A be a set of all 4-digit natural numbers whose exactly one digit is 7. Then the probability that a randomly chosen element of A leaves remainder 2 when divided by 5 is :

A

{2 \over 9}

B

{1 \over 5}

C

{122 \over 297}

D

{97 \over 297}

Correct Answer

Option D

Solution

n(s) = n(when 7 appears on thousands place) + n (7 does not appear on thousands place) = 9 $\times$ 9 $\times$ 9 + 8 $\times$ 9 $\times$ 9 $\times$ 3 = 33 $\times$ 9 $\times$ 9 n(E) = n(last digit 7 & 7 appears once) +n(last digit 2 when 7 appears once) = 8 $\times$ 9 $\times$ 9 + (3 $\times$ 9 $\times$ 9 - 2 $\times$ 9) $\therefore$

P(E) = {{8 \times 9 \times 9 + 9 \times 25} \over {33 \times 9 \times 9}} = {{97} \over {297}}

Q56

In a group of 400 people, 160 are smokers and non-vegetarian; 100 are smokers and vegetarian and the remaining 140 are non-smokers and vegetarian. Their chances of getting a particular chest disorder are 35%, 20% and 10% respectively. A person is chosen from the group at random and is found to be suffering from the chest disorder. The probability that the selected person is a smoker and non-vegetarian is :

A

{{14} \over {45}}

B

{{8} \over {45}}

C

{{7} \over {45}}

D

{{28} \over {45}}

Correct Answer

Option D

Solution

Consider following events A : Person chosen is a smoker and non vegetarian.

B : Person chosen in a smoker and vegetarian.

C : Person chosen is a non-smoker and vegetarian.

E : Person chosen has a chest disorder Given

P(A) = {{160} \over {400}}

P(B) = {{100} \over {400}}

P(C) = {{140} \over {400}}

P\left( {{E \over A}} \right) = {{35} \over {100}}

P\left( {{E \over B}} \right) = {{20} \over {100}}

P\left( {{E \over C}} \right) = {{10} \over {100}}

To find

P\left( {{A \over E}} \right) = {{P(A)P\left( {{E \over A}} \right)} \over {P(A).P\left( {{E \over A}} \right) + P(B).P\left( {{E \over B}} \right) + P(C).P\left( {{E \over C}} \right)}}

= {{{{160} \over {400}} \times {{35} \over {100}}} \over {{{160} \over {400}} \times {{35} \over {100}} \times {{100} \over {400}} \times {{20} \over {200}} + {{140} \over {400}} \times {{10} \over {100}}}}

= {{28} \over {45}}

Q57

A seven digit number is formed using digits 3, 3, 4, 4, 4, 5, 5. The probability, that number so formed is divisible by 2, is :

A

{1 \over 7}

B

{4 \over 7}

C

{6 \over 7}

D

{3 \over 7}

Correct Answer

Option D

Solution

Digits = 3, 3, 4, 4, 4, 5, 5 Total 7 digit numbers =

{{7!} \over {2!2!3!}}

Number of 7 digit number divisible by 2 $\Rightarrow$ last digit = 4 Now 7 digit numbers which are divisible by 2 =

{{6!} \over {2!2!2!}}

Required probability =

{{{{6!} \over {2!2!2!}}} \over {{{7!} \over {3!2!2!}}}} = {3 \over 7}

Q58

Let A denote the event that a 6-digit integer formed by 0, 1, 2, 3, 4, 5, 6 without repetitions, be divisible by 3. Then probability of event A is equal to :

A

{4 \over {9}}

B

{9 \over {56}}

C

{11 \over {27}}

D

{3 \over {7}}

Correct Answer

Option A

Solution

Total cases :

\underline 6

.

\underline 6

.

\underline 5

.

\underline 4

.

\underline 3

.

\underline 2

n(s) = 6 .

6!

Favourable cases : Number divisible by 3 $\equiv$ Sum of digits must be divisible by 3 Case - I 1, 2, 3, 4, 5, 6 Number of ways = 6!

Case - II 0, 1, 2, 4, 5, 6 Number of ways = 5 .

5!

Case - III 0, 1, 2, 3, 4, 5 Number of ways = 5 .

5!

n(favourable) = 6!

+ 2 .

5 .

5!

P = {{6! + 2.\,5.\,5!} \over {6\,.\,6!}} = {4 \over 9}

Q59

Two dies are rolled. If both dices have six faces numbered 1, 2, 3, 5, 7 and 11, then the probability that the sum of the numbers on the top faces is less than or equal to 8 is :

A

{4 \over 9}

B

{1 \over 2}

C

{5 \over {12}}

D

{17 \over {36}}

Correct Answer

Option D

Solution

n(S) = 36 possible ordered pair : (1, 1), (1, 2), (1, 3), (1, 5), (1, 7), (2, 1), (2, 2), (2, 3), (2, 5), (3, 1), (3, 2), (3, 3), (3, 5), (5, 1), (5, 2), (5, 3), (7, 1) Number of favourable outcomes = 17 Probability =

{{17} \over {36}}

Q60

Let a computer program generate only the digits 0 and 1 to form a string of binary numbers with probability of occurrence of 0 at even places be

{1 \over 2}

and probability of occurrence of 0 at the odd place be

{1 \over 3}

. Then the probability that '10' is followed by '01' is equal to :

A

{1 \over 18}

B

{1 \over 3}

C

{1 \over 9}

D

{1 \over 6}

Correct Answer

Option C

Solution

P(0 at even place) =

{1 \over 2}

P(0 at odd place) =

{1 \over 3}

P(1 at even place) =

{1 \over 2}

P(1 at odd place) =

{2 \over 3}

P(10 is followed by 01) =

\left( {{2 \over 3} \times {1 \over 2} \times {1 \over 3} \times {1 \over 2}} \right) + \left( {{1 \over 2} \times {1 \over 3} \times {1 \over 2} \times {2 \over 3}} \right)

=

{1 \over {18}} + {1 \over {18}}

=

{1 \over 9}