5. A standard deck of cards is well shuffled. Let AC denote the ace of clubs and

Question

5. A standard deck of cards is well shuffled. Let AC denote the ace of clubs and let KD denote the king of diamonds. Explain your answer. Think before calculating. (a) What is the probability the top card is AC? (b) What is the probability the second card is AC? (c) What is the probability the second card is AC given the top card is AC? (d) What is the probability the second card is AC given the top card is KD? e) What is the probability the 5th card is AC? (f) What is the probability the 5th card is AC given the second card is KD?

Explanation / Answer

a) The probability the top card is AC i.e. P(AC ) = 1/52

b) the probabilty that second card is AC i.e. P(AC in Second card) = P(AC is not came in first card) * P(AC is came in Second card)

= 51/52 * 1/51 = 1/52

c) The probabilty the second card is AC given the top card is AC is

P(second card is AC / top card is AC) = P(second card is AC and top card is AC) / P(top card is AC)

=0 / (1/52) = 0

d)

The probabilty the second card is AC given the top card is KD is

P(second card is AC / top card is KD) = P(second card is AC and top card is KD) / P(top card is KD)

=(1/52)(1/51) / (1/52) = 1/51

e) The probabilty the 5th card is AC = P(1st 4 card are not drawn AC and 5th time is drawn AC)

= (51/52)*(50/51)(49/50)(48/49)(1/48) = 1/52

f) P(5th card is AC / Second card is KD) = P(5th card is AC and Second card is KD) / P(second card is KD)

= [ (50/52) (1/51) (49/50) (48/49) (1/48) ] / [(51/52) * (1/51)] = (50*49*48) / (52*50*49*48) = 1/52

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