Two cards are drawn simultaneously from a pack of 52 cards find the probability distribution of aces

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Let X denote the number of aces in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,

\[P\left( X = 0 \right)\]

\[ = P\left( \text{ no ace } \right)\]

\[ = \frac{48}{52} \times \frac{48}{52}\]

\[ = \frac{12 \times 12}{13 \times 13}\]

\[ = \frac{144}{169}\]

\[P\left( X = 1 \right)\]

\[ = P\left( 1 \text{ ace }  \right)\]

\[ = \frac{4}{52} \times \frac{48}{52}\]

\[ = \frac{2 \times 12}{13 \times 13}\]

\[ = \frac{24}{169}\]

\[P\left( X = 2 \right)\]

\[ = P\left( 2 \text{ aces }  \right)\]

\[ = \frac{4}{52} \times \frac{4}{52}\]

\[ = \frac{1 \times 1}{13 \times 13}\]

\[ = \frac{1}{169}\]

Thus, the probability distribution of X is given by

Page 2

Let X denote the number of kings in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,

\[P\left( X = 0 \right)\]

\[ = P\left( \text{ no kings } \right)\]

\[ = \frac{48}{52} \times \frac{48}{52}\]

\[ = \frac{12 \times 12}{13 \times 13}\]

\[ = \frac{144}{169}\]

\[P\left( X = 1 \right)\]

\[ = P\left( 1 \text{ king } \right)\]

\[ = \frac{4}{52} \times \frac{48}{52}\]

\[ = \frac{2 \times 12}{13 \times 13}\]

\[ = \frac{24}{169}\]

\[P\left( X = 2 \right)\]

\[ = P\left( 2 \text{ kings } \right)\]

\[ = \frac{4}{52} \times \frac{4}{52}\]

\[ = \frac{1 \times 1}{13 \times 13}\]

\[ = \frac{1}{169}\]

Thus, the probability distribution of X is given by

Page 3

Let X denote the number of aces in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,

\[P\left( X = 0 \right)\]

\[ = P\left( \text{ no ace } \right)\]

\[ = \frac{48}{52} \times \frac{47}{51}\]

\[ = \frac{2256}{2652}\]

\[ = \frac{188}{221}\]

\[P\left( X = 1 \right)\]

\[ = P\left( 1 \text{ ace } \right)\]

\[ = \frac{4}{52} \times \frac{48}{51} + \frac{48}{52} \times \frac{4}{51}\]

\[ = \frac{384}{2652}\]

\[ = \frac{32}{221}\]

\[P\left( X = 2 \right)\]

\[ = P\left( 2 \text{ aces } \right)\]

\[ = \frac{4}{52} \times \frac{3}{51}\]

\[ = \frac{12}{2652}\]

\[ = \frac{1}{221}\]

Thus, the probability distribution of X is given by