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08 Feb 2021, 03:00
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B
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Difficulty:
95%
(hard)
Question Stats:
45% (02:33) correct
55%
(02:44)
wrong
based on 105
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Two cards are drawn at random from a pack of 52 cards without replacement. What is the probability that one is a club and other is an ace?
A. 1/52
B. 1/26
C. 1/13
D. 4/13
E. 7/13
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
A. 1/52
B. 1/26
C. 1/13
D. 4/13
E. 7/13
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
08 Feb 2021, 05:26
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Alternate approach:
AND means MULTIPLY.
OR means ADD.
Case 1: P(ace of spades, diamonds or hearts AND any type of clubs)
P(first card is an ace of spades, diamonds or hearts) = 3/52 (Of the 52 cards, 3 are the ace of spades, diamonds or hearts)
P(second card is clubs) = 13/51 (Of the 51 remaining cards, 13 are clubs)
Since Case 1 requires the first card AND the second card, we multiply:
3/52 * 13/51 = 1/4 * 1/17
Since the two cards can also be picked in reverse order, we multiply by 2:
1/4 * 1/17 * 2 = 1/2 * 1/17
Case 2: P(ace of clubs AND any other type of clubs)
P(first card is ace of clubs) = 1/52 (Of the 52 cards, 1 is the ace of clubs)
P(second card is clubs) = 12/51 (Of the 51 remaining cards, 12 are clubs)
Since Case 2 requires the first card AND the second card, we multiply:
1/52 * 12/51 = 1/13 * 1/17
Since the two cards can also be picked in reverse order, we multiply by 2:
1/13 * 1/17 * 2 = 2/13 * 1/17
Since a favorable outcome will be yielded by Case 1 OR Case 2, we add the results in blue:
(1/2 * 1/17) + (2/13 +1/17) = (1/17)(1/2 + 2/13) = (1/17)(17/26) = 1/26
Bunuel
AND means MULTIPLY.
OR means ADD.
Case 1: P(ace of spades, diamonds or hearts AND any type of clubs)
P(first card is an ace of spades, diamonds or hearts) = 3/52 (Of the 52 cards, 3 are the ace of spades, diamonds or hearts)
P(second card is clubs) = 13/51 (Of the 51 remaining cards, 13 are clubs)
Since Case 1 requires the first card AND the second card, we multiply:
3/52 * 13/51 = 1/4 * 1/17
Since the two cards can also be picked in reverse order, we multiply by 2:
1/4 * 1/17 * 2 = 1/2 * 1/17
Case 2: P(ace of clubs AND any other type of clubs)
P(first card is ace of clubs) = 1/52 (Of the 52 cards, 1 is the ace of clubs)
P(second card is clubs) = 12/51 (Of the 51 remaining cards, 12 are clubs)
Since Case 2 requires the first card AND the second card, we multiply:
1/52 * 12/51 = 1/13 * 1/17
Since the two cards can also be picked in reverse order, we multiply by 2:
1/13 * 1/17 * 2 = 2/13 * 1/17
Since a favorable outcome will be yielded by Case 1 OR Case 2, we add the results in blue:
(1/2 * 1/17) + (2/13 +1/17) = (1/17)(1/2 + 2/13) = (1/17)(17/26) = 1/26
General Discussion
08 Feb 2021, 04:27
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Bunuel
We have 13 clubs cards, out of which one is an Ace.
Total ways to select 2 cards out of 52 = \(52C2=52*\frac{51}{2}=26*51\)
Ways to select the given scenario
1) Any of 3 aces and one of the 13 clubs(including the club ace) = \(3C1*13C1=3*13=39\)
2) Ace of club and one out of the remaining 12 club cards = \(1C1*12C1=1*12=12\)
Total = 39+12=51
\(P= \frac{51}{26*51}=\frac{1}{26}\)
B
