In a certain game, you pick a card from a standard deck of 52 cards.
In a certain game, you pick a card from a standard deck of 52 cards. If the card is a heart, you win. If the card is not a heart, the person replaces the card to the deck, reshuffles, and draws again. The person keeps repeating that process until he picks a heart, and the point is to measure: how many draws did it take before the person picked a heart and won? What is the probability that one will have at least two “heartless” draws on the first two draws, not picking the first heart until at least the third draw?
Answer/Solution
9/16
Steps/Work
A full deck of 52 cards contains 13 cards from each of the four suits. The probability of drawing a heart from a full deck is 1/4. Therefore, the probability of “not heart” is 3/4.
P(at least three draws to win) = 1 – P(win in two or fewer draws)
Furthermore,
P(win in two or fewer draws) = P(win in one draw OR win in two draws)
= P(win in one draw) + P(win in two draws)
Winning in one draw means: I select one card from a full deck, and it turns out to be a heart. Above, we already said: the probability of this is 1/4.
P(win in one draw) = 1/4
Winning in two draws means: my first draw is “not heart”, P = 3/4, AND the second draw is a heart, P = 1/4. Because we replace and re-shuffle, the draws are independent, so the AND means multiply.
P(win in two draws) =(3/4)*(1/4) = 3/16
P(win in two or fewer draws) =P(win in one draw) + P(win in two draws)
= 1/4 + 3/16 = 7/16
P(at least three draws to win) = 1 – P(win in two or fewer draws)
= 1 – 7/16 = 9/16
Answer = B
P(at least three draws to win) = 1 – P(win in two or fewer draws)
Furthermore,
P(win in two or fewer draws) = P(win in one draw OR win in two draws)
= P(win in one draw) + P(win in two draws)
Winning in one draw means: I select one card from a full deck, and it turns out to be a heart. Above, we already said: the probability of this is 1/4.
P(win in one draw) = 1/4
Winning in two draws means: my first draw is “not heart”, P = 3/4, AND the second draw is a heart, P = 1/4. Because we replace and re-shuffle, the draws are independent, so the AND means multiply.
P(win in two draws) =(3/4)*(1/4) = 3/16
P(win in two or fewer draws) =P(win in one draw) + P(win in two draws)
= 1/4 + 3/16 = 7/16
P(at least three draws to win) = 1 – P(win in two or fewer draws)
= 1 – 7/16 = 9/16
Answer = B