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 · 27 Aug · Shift 2 · Q36
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 · 27 Aug · Shift 2 · Q36

Probability question

2021 · 27 Aug · Shift 2 · Q36

JEE MainMathematicsProbabilityNumerical+4 / −1
The probability distribution of random variable X is given by :

X 1 2 3 4 5
P(X) K 2K 2K 3K K


Let p = P(1 < X < 4 | X < 3). If 5p = λ\lambdaλ K, then λ\lambdaλ equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 30

  1. Write the distribution clearly

The random variable XXX takes values 1,2,3,4,51,2,3,4,51,2,3,4,5 with probabilities

P(X=1)=K,P(X=2)=2K,P(X=3)=2K,P(X=4)=3K,P(X=5)=K.P(X=1)=K,\quad P(X=2)=2K,\quad P(X=3)=2K,\quad P(X=4)=3K,\quad P(X=5)=K.P(X=1)=K,P(X=2)=2K,P(X=3)=2K,P(X=4)=3K,P(X=5)=K.
  1. Use total probability = 1 to find KKK
K+2K+2K+3K+K=9K=1K+2K+2K+3K+K = 9K = 1K+2K+2K+3K+K=9K=1

So,

K=19.K=\frac{1}{9}.K=91​.
  1. Compute p=P(1<X<4∣X<3)p=P(1<X<4\mid X<3)p=P(1<X<4∣X<3)

By definition of conditional probability,

p=P((1<X<4)∩(X<3))P(X<3).p=\frac{P((1<X<4)\cap (X<3))}{P(X<3)}.p=P(X<3)P((1<X<4)∩(X<3))​.

Now,

  • 1<X<41<X<41<X<4 means X=2X=2X=2 or X=3X=3X=3.
  • X<3X<3X<3 means X=1X=1X=1 or X=2X=2X=2.

Their intersection is only X=2X=2X=2.

Hence,

P((1<X<4)∩(X<3))=P(X=2)=2K.P((1<X<4)\cap (X<3))=P(X=2)=2K.P((1<X<4)∩(X<3))=P(X=2)=2K.

Also,

P(X<3)=P(X=1)+P(X=2)=K+2K=3K.P(X<3)=P(X=1)+P(X=2)=K+2K=3K.P(X<3)=P(X=1)+P(X=2)=K+2K=3K.

Therefore,

p=2K3K=23.p=\frac{2K}{3K}=\frac{2}{3}.p=3K2K​=32​.
  1. Use the given relation 5p=λK5p=\lambda K5p=λK

Substitute p=23p=\frac{2}{3}p=32​ and K=19K=\frac{1}{9}K=91​:

5p=5⋅23=103.5p=5\cdot \frac{2}{3}=\frac{10}{3}.5p=5⋅32​=310​.

So,

λK=103.\lambda K = \frac{10}{3}.λK=310​.

Since K=19K=\frac{1}{9}K=91​,

λ⋅19=103\lambda\cdot \frac{1}{9} = \frac{10}{3}λ⋅91​=310​ λ=103⋅9=30.\lambda = \frac{10}{3}\cdot 9 = 30.λ=310​⋅9=30.
  1. Final answer
30\boxed{30}30​
PreviousNext

More from Probability

  • The probability that a randomly selected 2-digit number belongs to the set {n ∈ N : (2n − 2) is a multiple of 3} is equal to :2021 · MCQ
  • An electric instrument consists of two units. Each unit must function independently for the instrument to operate. The probability that the first unit functions is 0.9 and that of the second unit is 0.8. The instrument is switched on and…2021 · Numerical
  • 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 :2021 · MCQ
  • Box I contains 30 cards numbered 1 to 30 and Box II contains 20 cards numbered 31 to 50. A box is selected at random and a card is drawn from it. The number on the card is found to be a non-prime number. The probability that the card was…2020 · MCQ
  • Let EC denote the complement of an event E. Let E1 , E2 and E3 be any pairwise independent events with P(E1) > 0 and P(E1 ∩ E2 ∩ E3) = 0. Then P(E2C​∩E3C​/E1​) is equal to :2020 · MCQ
  • A dice is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of 4. Then the conditional probability that the score 4 has appeared atleast once is :2020 · MCQ
  • The probability that a randomly chosen 5-digit number is made from exactly two digits is :2020 · MCQ
  • The probability of a man hitting a target is 101​. The least number of shots required, so that the probability of his hitting the target at least once is greater than 41​, is ​.2020 · Numerical