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From a pack of 52 cards, the cards are drawn one by one till an ace appears. The chance that an ace does not come up in first 26 cards is
- a)109/153
- b)23/27
- c)46/153
- d)none of these
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
From a pack of 52 cards, the cards are drawn one by one till an ace ap...
There are 4 aces in a pack. So, the probability of drawing an ace = 4/52
= 1/13
and the probability of drawing a card other than ace = 12/13.
Hence, the probability that ace doesn't appear in first 26 draws = (12/13)26
= 1/13
and the probability of drawing a card other than ace = 12/13.
Hence, the probability that ace doesn't appear in first 26 draws = (12/13)26
Most Upvoted Answer
From a pack of 52 cards, the cards are drawn one by one till an ace ap...
Solution:
Let the probability that an ace appears in the first draw be P(A).
Then, the probability that an ace does not appear in the first draw is P(A') = 1 - P(A).
Now, for the second draw, there are 51 cards left, out of which there are 3 aces left.
So, the probability that an ace appears in the second draw, given that an ace did not appear in the first draw, is P(B|A') = 3/51.
Similarly, for the third draw, there are 50 cards left, out of which there are 2 aces left.
So, the probability that an ace appears in the third draw, given that aces did not appear in the first two draws, is P(C|A' ∩ B') = 2/50.
And for the fourth draw, there are 49 cards left, out of which there is only 1 ace left.
So, the probability that an ace appears in the fourth draw, given that aces did not appear in the first three draws, is P(D|A' ∩ B' ∩ C') = 1/49.
Now, the probability that an ace does not appear in the first 26 cards is given by:
P(ace does not appear in first 26 cards) = P(A') × P(B'|A') × P(C'|A' ∩ B') × P(D'|A' ∩ B' ∩ C')
= (1 - P(A)) × (48/51) × (48/50) × (48/49)
= (1/13) × (16/17) × (24/25) × (48/49)
= 3072/270725
= 0.01134
Therefore, the correct answer is option D, none of these.
Let the probability that an ace appears in the first draw be P(A).
Then, the probability that an ace does not appear in the first draw is P(A') = 1 - P(A).
Now, for the second draw, there are 51 cards left, out of which there are 3 aces left.
So, the probability that an ace appears in the second draw, given that an ace did not appear in the first draw, is P(B|A') = 3/51.
Similarly, for the third draw, there are 50 cards left, out of which there are 2 aces left.
So, the probability that an ace appears in the third draw, given that aces did not appear in the first two draws, is P(C|A' ∩ B') = 2/50.
And for the fourth draw, there are 49 cards left, out of which there is only 1 ace left.
So, the probability that an ace appears in the fourth draw, given that aces did not appear in the first three draws, is P(D|A' ∩ B' ∩ C') = 1/49.
Now, the probability that an ace does not appear in the first 26 cards is given by:
P(ace does not appear in first 26 cards) = P(A') × P(B'|A') × P(C'|A' ∩ B') × P(D'|A' ∩ B' ∩ C')
= (1 - P(A)) × (48/51) × (48/50) × (48/49)
= (1/13) × (16/17) × (24/25) × (48/49)
= 3072/270725
= 0.01134
Therefore, the correct answer is option D, none of these.
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From a pack of 52 cards, the cards are drawn one by one till an ace appears. The chance that an ace does not come up in first 26 cards isa)109/153b)23/27c)46/153d)none of theseCorrect answer is option 'D'. Can you explain this answer?
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