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Probability question

2020 · 9 Jan · Shift 2 · Q34
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Probability question

2020 · 9 Jan · Shift 2 · Q34

JEE MainMathematicsProbabilityMCQ+4 / −1
A random variable X has the following probability distribution :

X: 1 2 3 4 5
P(X): K2 2K K 2K 5K2

Then P(X > 2) is equal to :
  1. A
    16{1 \over {6}}61​
  2. B
    712{7 \over {12}}127​
  3. C
    136{1 \over {36}}361​
  4. D
    2336{23 \over {36}}3623​
View written solutionFree

Correct answer: D

  1. Write the given distribution

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

P(X=1)=K2, P(X=2)=2K, P(X=3)=K, P(X=4)=2K, P(X=5)=5K2.P(X=1)=K^2,\, P(X=2)=2K,\, P(X=3)=K,\, P(X=4)=2K,\, P(X=5)=5K^2.P(X=1)=K2,P(X=2)=2K,P(X=3)=K,P(X=4)=2K,P(X=5)=5K2.

  1. Use the fact that total probability is 1

Since these form a probability distribution,

K2+2K+K+2K+5K2=1.K^2+2K+K+2K+5K^2=1.K2+2K+K+2K+5K2=1.

So,

6K2+5K=16K^2+5K=16K2+5K=1

6K2+5K−1=0.6K^2+5K-1=0.6K2+5K−1=0.

  1. Solve for KKK

Factorizing,

6K2+5K−1=(6K−1)(K+1)=0.6K^2+5K-1=(6K-1)(K+1)=0.6K2+5K−1=(6K−1)(K+1)=0.

Hence,

K=16orK=−1.K=\frac{1}{6} \quad \text{or} \quad K=-1.K=61​orK=−1.

Since probabilities cannot be negative, we reject K=−1K=-1K=−1.

Thus,

K=16.K=\frac{1}{6}.K=61​.

  1. Find P(X>2)P(X>2)P(X>2)

Now,

P(X>2)=P(X=3)+P(X=4)+P(X=5).P(X>2)=P(X=3)+P(X=4)+P(X=5).P(X>2)=P(X=3)+P(X=4)+P(X=5).

Substitute the values:

P(X>2)=K+2K+5K2=3K+5K2.P(X>2)=K+2K+5K^2=3K+5K^2.P(X>2)=K+2K+5K2=3K+5K2.

Using K=16K=\frac{1}{6}K=61​,

P(X>2)=3(16)+5(16)2P(X>2)=3\left(\frac{1}{6}\right)+5\left(\frac{1}{6}\right)^2P(X>2)=3(61​)+5(61​)2

=36+536=\frac{3}{6}+\frac{5}{36}=63​+365​

=12+536=\frac{1}{2}+\frac{5}{36}=21​+365​

=1836+536=\frac{18}{36}+\frac{5}{36}=3618​+365​

=2336.=\frac{23}{36}.=3623​.

  1. Check options

2336\frac{23}{36}3623​ corresponds to Option D.

Final Answer: 2336\boxed{\frac{23}{36}}3623​​

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