Question

Two bad eggs are accidently mixed up with ten good ones. Three eggs are drawn at random with replacement from this lot. Compute the mean for the number of bad eggs drawn.

Answer 1

As there are total of two bad eggs. Therefore while drawing 3 eggs we can draw 1 bad egg or 2 or 0 bad eggs.

Let X be the random variable denoting number of bad eggs that can be drawn in each draw.

Clearly X can take values 0, 1 or 2

P(X = 0) = P(all 3 are good eggs)

[Since there are 10 good eggs so for selecting all good we took all three from 10 and 0 eggs from 2 bad ones. Total sample points are no of ways of selecting 3 eggs from total of 12 eggs]

Similarly,

P(X = 1) = P(1 bad and 2 good eggs)

P(X = 2) = P(2 Bad eggs and 1 good egg)

Now we have p_i and x_i.

Let’s proceed to find mean

Mean of any probability distribution is given by Mean = ∑x_ip_i

∴ first we need to find the products i.e. p_ix_i and add them to get mean.

Following table gives the required products :