Math, asked by promeelab, 22 days ago

5 cards are drawn successively from a well-shuffled pack of 52 cards with replacement. Determine the probability that (i) all the five cards should be spades? (ii) only 3 cards should be spades? (iii) none of the cards is a spade?​

Answers

Answered by sugantipandit601
0

(i) P(all cards are spade) = 5C5 (+)³ () º 0

(i) P(all cards are spade) = 5C5 (+)³ () º 05 - (+)³

(i) P(all cards are spade) = 5C5 (+)³ () º 05 - (+)³1 1024

(i) P(all cards are spade) = 5C5 (+)³ () º 05 - (+)³1 1024(ii) P(only three cards are spade) = 5C3 ³(-)³ (3) ²

(i) P(all cards are spade) = 5C5 (+)³ () º 05 - (+)³1 1024(ii) P(only three cards are spade) = 5C3 ³(-)³ (3) ²51 31 21 9 1024 =

(i) P(all cards are spade) = 5C5 (+)³ () º 05 - (+)³1 1024(ii) P(only three cards are spade) = 5C3 ³(-)³ (3) ²51 31 21 9 1024 =45 512 =

(i) P(all cards are spade) = 5C5 (+)³ () º 05 - (+)³1 1024(ii) P(only three cards are spade) = 5C3 ³(-)³ (3) ²51 31 21 9 1024 =45 512 =(iii) P(none of them are spade) = 5Co

(i) P(all cards are spade) = 5C5 (+)³ () º 05 - (+)³1 1024(ii) P(only three cards are spade) = 5C3 ³(-)³ (3) ²51 31 21 9 1024 =45 512 =(iii) P(none of them are spade) = 5Co-

(i) P(all cards are spade) = 5C5 (+)³ () º 05 - (+)³1 1024(ii) P(only three cards are spade) = 5C3 ³(-)³ (3) ²51 31 21 9 1024 =45 512 =(iii) P(none of them are spade) = 5Co-243 1024

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