Question:
A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The probability that the missing card is not a spade, is :
A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The probability that the missing card is not a spade, is :
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Use Bayes' theorem with prior \(\frac14\) (missing card is a spade) and \(\frac34\) (not a spade), and compute the chance that two drawn cards are both spades in each case.
Updated On: Nov 4, 2026
- \(\frac{39}{50}\)
- \(\frac{3}{4}\)
- \(\frac{22}{425}\)
- \(\frac{52}{867}\)
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The Correct Option is A
Solution and Explanation
Let E: two spades drawn from the 51 cards; \(M_s\): missing card is a spade (prob \(\frac14\)); \(M_n\): missing card is not a spade (prob \(\frac34\)).
\(P(E|M_s)=\frac{\binom{12}{2}}{\binom{51}{2}}=\frac{66}{1275}\), \(P(E|M_n)=\frac{\binom{13}{2}}{\binom{51}{2}}=\frac{78}{1275}\).
\(P(M_n|E)=\frac{\frac34\cdot78}{\frac14\cdot66+\frac34\cdot78}=\frac{234}{66+234}=\frac{234}{300}=\frac{39}{50}\).
Answer: (1) \(\frac{39}{50}\)
\(P(E|M_s)=\frac{\binom{12}{2}}{\binom{51}{2}}=\frac{66}{1275}\), \(P(E|M_n)=\frac{\binom{13}{2}}{\binom{51}{2}}=\frac{78}{1275}\).
\(P(M_n|E)=\frac{\frac34\cdot78}{\frac14\cdot66+\frac34\cdot78}=\frac{234}{66+234}=\frac{234}{300}=\frac{39}{50}\).
Answer: (1) \(\frac{39}{50}\)
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