Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Probability question

2021 · 24 Feb · Shift 2 · Q31
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Probability
  5. /2021 · 24 Feb · Shift 2 · Q31

Probability question

2021 · 24 Feb · Shift 2 · Q31

JEE MainMathematicsProbabilityMCQ+4 / −1
The probability that two randomly selected subsets of the set {1, 2, 3, 4, 5} have exactly two elements in their intersection, is :
  1. A
    13529{{135} \over {{2^9}}}29135​
  2. B
    6528{{65} \over {{2^8}}}2865​
  3. C
    6527{{65} \over {{2^7}}}2765​
  4. D
    3527{{35} \over {{2^7}}}2735​
View written solutionFree

Correct answer: A

  1. Total number of ordered pairs of subsets

Let the two randomly selected subsets be AAA and BBB of the set S={1,2,3,4,5}.S=\{1,2,3,4,5\}.S={1,2,3,4,5}.

A set with 5 elements has 25=322^5=3225=32 subsets.

So the total number of ordered pairs (A,B)(A,B)(A,B) is 32×32=25⋅25=210=1024.32\times 32=2^5\cdot 2^5=2^{10}=1024.32×32=25⋅25=210=1024.

  1. Count pairs with exactly two common elements

We want ∣A∩B∣=2.|A\cap B|=2.∣A∩B∣=2.

First choose the 2 elements that will be in the intersection: (52)=10.\binom{5}{2}=10.(25​)=10.

Now consider the remaining 333 elements. For each such element, to keep the intersection exactly 2, it cannot be in both AAA and BBB simultaneously.

So for each of these 3 elements, the allowed possibilities are:

  • in neither set,
  • in AAA only,
  • in BBB only.

Thus each remaining element has 333 choices, giving 33=273^3=2733=27 ways.

Hence favorable ordered pairs are (52)⋅33=10⋅27=270.\binom{5}{2} \cdot 3^3 = 10\cdot 27 = 270.(25​)⋅33=10⋅27=270.

  1. Compute probability

Therefore, P(∣A∩B∣=2)=2701024.P(|A\cap B|=2)=\frac{270}{1024}.P(∣A∩B∣=2)=1024270​.

Simplify: 2701024=135512=13529.\frac{270}{1024}=\frac{135}{512}=\frac{135}{2^9}.1024270​=512135​=29135​.

  1. Match with options

13529\frac{135}{2^9}29135​ corresponds to Option A.

  1. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.

PreviousNext

More from Probability

  • When a missile is fired from a ship, the probability that it is intercepted is 31​ and the probability that the missile hits the target, given that it is not intercepted, is 43​. If three missiles are fired…2021 · MCQ
  • 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 :2021 · MCQ
  • 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 :2021 · MCQ
  • 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…2021 · MCQ
  • 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(43​)9 then k lies in the set :2021 · MCQ
  • Let X be a random variable such that the probability function of a distribution is given by P(X=0)=21​,P(X=j)=3j1​(j=1,2,3,...,∞). Then the mean of the distribution and P(X is positive and even)…2021 · MCQ
  • A fair coin is tossed n-times such that the probability of getting at least one head is at least 0.9. Then the minimum value of n is ​.2021 · Numerical
  • 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) = 95​, is :2021 · MCQ