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the question and the solution is in the attachment, can you explain how can they use combinations when combinations does not account for replacement while it is mentioned in the question that replacement is to be taken into account? And can you provide the correct answer.
Asked by ashwinskrishna2006 | 26 Dec, 2023, 07:34: PM

You are correct. The solution given in the posted image is wrong.

For example , the solution gives the probability p(0) of getting no spades as

..............................................(1)

above expression for probability p(0) of getting spades is wrong.

Denominator of above expression (1) is number of ways of randomly removing two cards from pack of 52 cards once .

But we need number of ways of removing two cards randomly one after other by replacing the first card before removing second time.

Number of ways of removing two cards randomly one after other by replacemet is  52C1 × 52C1 .

Numerator of above expression (1) is number of ways of randomly removing two non-spade cards from 39 cards of non-spade .

Number of ways of removing two non-spade cards randomly one after other by replacemet is  39C1 × 39C1 .

Hence probability of getting no spades if two cards removed one after another by replacement is

Probability of getting one spade if two cards removed one after another by replacement is

Probability of getting two spades if two cards removed one after another by replacement is

Answered by Thiyagarajan K | 27 Dec, 2023, 12:17: PM

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