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Hi Wei Ren

N88/II/6

a) P(all the four cards are different) = (10x9x8x7)/(10x10x10x10) = 63/125

b) Since the first three are different. He has 10x9x8 choices. The fourth card is the same as either one of the first 3 cards, therefore he has 3 choices. Althogether he has 10x9x8x3 choices.

P(exactly 3 different pictures out of a total of four cards) = (10x9x8x3)/10^4 = 27/125

c) Required probability = 1 – P(all the four cards does not contain X or Y)= 1- (8/10)^4 = 0.591

d) P(n more cards are needed) (9/10)^n-1 x (1/10)

P( at most n more cards are needed) = 1/10 + (9/10)(1/10)………..(9/10)^n-1 x (1/10)

This is sum of GP

1- (9/10)^n > 099

n > 43.7

least value n = 44

From A level/H2 Maths Tutor