Practicing Success
A draws two cards at random from a pack of 52 cards. After returning them to the pack and shuffling it, B draws two cards at random. The probability that their draws contain exactly one common cards is |
$\frac{25}{546}$ $\frac{50}{663}$ $\frac{25}{663}$ none of these |
$\frac{50}{663}$ |
Let us number the cards as 1, 2, 3, ...., 52. Also, let $A_i(i= 1, 2, ....52)$ denote the event $i^{th}$ card is common in the two draws of two cards each. Then, Required probability $= P(A_1 ∪ A_1 ∪ ....∪ A_{52})$ $=\sum\limits^{52}_{r=1}P(A_r)$ $=\sum\limits^{52}_{r=1}P(A_r)$ Probability that the cards marked r is common to two draws of two cards each. $=\sum\limits^{52}_{r=1}P(A_r)$ {Probability of a drawing the card marked r and any other card) × (Probability of B drawing the card make r and any other card) $=\sum\limits^{52}_{r=1}P(A_r) \left(\frac{^{51}C_1}{^{52}C_2} × \frac{^{50}C_1}{^{52}C_2}\right) = 52 × \frac{51}{51×26}×\frac{50}{51×26}=\frac{50}{663}$ |