Let us denote the number of defective bulbs by X. Clearly X can take the values 0, 1, 2, 3, 4.
P(X = 0) = (no defective bulb) = P(all 4 goods ones) = (20!/(4!*16!))/(25!/(21!*4!))=969/2530
P(X = 1) = P(1 defective and 3 good ones) = (5!/(4!*1!)*20!/(17!*3!))/(25!/(21!/4!))=114/253
P(X = 2) = P(2 defective and 2 good ones) = (5!/(3!*2!)*20!/(18!*2!))/(25!/(21!*4!))=38/253
P(X = 3) = P(3 defective and one good one) = (5!/(3!*2!)*20!/(19!*1!))/(25!/(21!*4!))=4/253
P(X = 4) = P(all 4 defective) = (5!/(4!*1!))/(25!/(21!*4!))=1/2530
So probability distribution is:
x p_i x*p_i
0 969/2530 0
1 114/253 114/253
2 38/253 76/253
3 4/253 12/253
4 1/2530 4/2530
The sum of all x*p_i equals 4/5.
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