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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 three draws before one picks a heart?
  • a)
    1/2
  • b)
    9/16
  • c)
    11/16
  • d)
    13/16
  • e)
    15/16
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
In a certain game, you pick a card from a standard deck of 52 cards. I...
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
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Most Upvoted Answer
In a certain game, you pick a card from a standard deck of 52 cards. I...
Calculating the Probability of Having At Least Three Draws Before Picking a Heart:
- The probability of not drawing a heart on the first draw is 39/52 (since there are 39 non-heart cards out of 52 total cards).
- The probability of not drawing a heart on the second draw is also 39/52.
- Therefore, the probability of not drawing a heart on the first two draws is (39/52) * (39/52).
- The probability of drawing a heart on the third draw is 13/52 (since there are 13 hearts in a deck of 52 cards).
- The probability of having at least three draws before picking a heart is equal to 1 minus the probability of picking a heart on the first or second draw, which is 1 - [(13/52) + (39/52)*(39/52)].
- Simplifying the equation, we get 1 - [13/52 + 1521/2704] = 1 - (676/2704) = 2028/2704 = 9/16.
Therefore, the probability that one will have at least three draws before picking a heart is 9/16, which corresponds to option 'b'.
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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 three draws before one picks a heart?a)1/2b)9/16c)11/16d)13/16e)15/16Correct answer is option 'B'. Can you explain this answer?
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