A card from a pack of 52 playing cards is lost. From the remaining cards of the pack three cards are drawn at random (without replacement) and are found to be all spades. Find the probability of the lost card being a spade.

OR

From a lot of 15 bulbs which include 5 defectives, a sample of 4 bulbs is drawn one by one with replacement. Find the probability distribution of number of defective bulbs. Hence find the mean of the distribution.

Let E₁ be the event that lost card is a spade.

⇒

Let E₂ be the event that lost card is not a spade.

⇒

Let A be the event drawing three spade cards from remaining cards without replacement.

Now,P(A/ E₁) = Probability of drawing three spade cards given that

lost card is spade.

⇒P(A/ E₁) = ¹²C₃ / ⁵¹C₃ =

P(A/ E₂) = Probability of drawing three spade cards given that

lost card is non spade.

⇒P(A/ E₂) = ¹³C₃ / ⁵¹C₃ =

By Baye’s theorem , we have required probability

=

⇒

OR

Let p = probability of getting defective bulb =

and q = probability of getting non defective bulb =

Let X be the number of defective bulbs out of 4 drawn.

X can take values 0,1,2,3 and 4.

Since X is a binomial variate with parameters n=4 , such that

P(X=r) = ⁿC_r.p^r.q^n-r when n=4 , r=0,1,2,3,4

P(X=0) = ⁴C₀.p⁰.q^4-0 = ⁴C₀

P(X=1) = ⁴C₁.p¹.q^4-1 = ⁴C₁

P(X=2) = ⁴C₂.p².q^4-2 = ⁴C₂

P(X=3) = ⁴C₃.p³.q^4-3 = ⁴C₃

P(X=4) = ⁴C₄.p⁴.q^4-4 = ⁴C₄

Thus, the probability distribution of X is given by

X: 0 1 2 3 4

P(x):

Now, mean E(X) =

=

=