Question

two cards are drawn from a well shuffled pack of 52 cards and thrown away. find the probability of drawing an ace from the remaining 50 cards

Answer 1

According to the total probability formula, we have

P(draw an ace from the remaining 50 cards) =

P(draw an ace | two drawn cards were aces) * P(draw two aces from 52 cards) +

P(draw an ace | two drawn cards were ace and not an ace) * P(draw one ace and one not ace from 52 cards) +

P(draw an ace | two drawn cards were not aces) * P(not to draw two aces from 52 cards).

Obviously,

P(draw an ace | two drawn cards were aces) = 0,

P(draw an ace | two drawn cards were ace and not an ace) = 3/50,

P(draw an ace | two drawn cards were not aces) = 4/50,

and

P(draw two aces from 52 cards) = 4/52 * 3/51,

P(draw one ace and one not ace from 52 cards) = 4/52 * (1 - 3/51) (or, equally, (1-4/52) * 4/51),

P(not to draw two aces from 52 cards) = (1 - 4/52)(1 - 4/51) (or, equally, 1 - 4/52 * 3/51).

So,

P(draw an ace from the remaining 50 cards) =

0 * 4/52 * 3/51 + 3/50 * 4/52 * (1 - 3/51) + 4/50 * (1 - 4/52)(1 - 4/51) =

= 3/50 * 4/52 * 48/51 + 4/50 * 48/52 * 47/51 = 9024/132600 = 1128/16575 ≈ 0.068.