Question

Two cards are drawn without replacement. The probability distribution of number of aces is given by:

• A.

  X	0	1	2
  P(X)	\(\frac{188}{221}\)	\(\frac{32}{221}\)	\(\frac{1}{221}\)

• B.

  X	0	1	2
  P(X)	\(\frac{144}{169}\)	\(\frac{24}{169}\)	\(\frac{1}{169}\)

• C.

  X	0	1	2
  P(X)	\(\frac{188}{221}\)	\(\frac{24}{221}\)	\(\frac{1}{221}\)

• D.

  X	0	1	2
  P(X)	\(\frac{188}{221}\)	\(\frac{1}{221}\)	\(\frac{24}{221}\)

Answer 1

The Correct Option is A

X	0	1	2
P(X)	\(\frac{188}{221}\)	\(\frac{32}{221}\)	\(\frac{1}{221}\)

To solve the problem, we need to verify which probability distribution matches the calculation based on the scenario given: drawing two cards without replacement from a deck and counting the number of aces drawn.

Consider a standard deck of 52 cards with 4 aces. The probabilities are calculated as follows:

• P(X=0): Probability of drawing 0 aces. Calculate the probability that both cards drawn are not aces. The number of non-aces is \(52-4=48\).

The probability that the first card is not an ace is \(\frac{48}{52}\), and for the second card (assuming the first was not an ace) is \(\frac{47}{51}\).

\[P(X=0)=\frac{48}{52} \times \frac{47}{51}=\frac{48 \times 47}{52 \times 51}=\frac{1888}{2652}=\frac{188}{221}\]

• P(X=1): Probability of drawing 1 ace and 1 non-ace. We have two cases: ace first or non-ace first.

1. Ace first: \(\frac{4}{52}\) (probability of ace) \(\times \frac{48}{51}\) (probability of non-ace).
2. Non-ace first: \(\frac{48}{52}\) (probability of non-ace) \(\times \frac{4}{51}\) (probability of ace).

\[P(X=1)=\left(\frac{4}{52} \times \frac{48}{51}\right)+\left(\frac{48}{52} \times \frac{4}{51}\right)\\=\frac{192}{2652}+\frac{192}{2652}=\frac{384}{2652}=\frac{32}{221}\]

• P(X=2): Probability of drawing 2 aces. First ace probability \(\frac{4}{52}\), second ace \(\frac{3}{51}\).

\[P(X=2)=\frac{4}{52} \times \frac{3}{51}=\frac{12}{2652}=\frac{1}{221}\]

The probabilities obtained match the values in the provided correct answer table.