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16 Sep 2014, 00:27
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If two cards are drawn without replacement from a standard 52-card deck, what is the probability that both are Jacks?
A. \(\frac{1}{13}\)
B. \(\frac{1}{17}\)
C. \(\frac{30}{221}\)
D. \(\frac{7}{221}\)
E. \(\frac{1}{221}\)
A. \(\frac{1}{13}\)
B. \(\frac{1}{17}\)
C. \(\frac{30}{221}\)
D. \(\frac{7}{221}\)
E. \(\frac{1}{221}\)
16 Sep 2014, 00:27
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Official Solution:
If two cards are drawn without replacement from a standard 52-card deck, what is the probability that both are Jacks?
A. \(\frac{1}{13}\)
B. \(\frac{1}{17}\)
C. \(\frac{30}{221}\)
D. \(\frac{7}{221}\)
E. \(\frac{1}{221}\)
The probability of drawing a Jack on the first draw is \(\frac{4}{52}\), since there are 4 Jacks in a 52-card deck. The probability of drawing a Jack on the second draw, after one Jack has already been taken, is \(\frac{3}{51}\). The overall probability of drawing both Jacks consecutively is the product of these two probabilities:
\(\frac{4}{52}*\frac{3}{51}= \frac{1}{13} * \frac{1}{17} = \frac{1}{221}\)
Answer: E
If two cards are drawn without replacement from a standard 52-card deck, what is the probability that both are Jacks?
A. \(\frac{1}{13}\)
B. \(\frac{1}{17}\)
C. \(\frac{30}{221}\)
D. \(\frac{7}{221}\)
E. \(\frac{1}{221}\)
The probability of drawing a Jack on the first draw is \(\frac{4}{52}\), since there are 4 Jacks in a 52-card deck. The probability of drawing a Jack on the second draw, after one Jack has already been taken, is \(\frac{3}{51}\). The overall probability of drawing both Jacks consecutively is the product of these two probabilities:
\(\frac{4}{52}*\frac{3}{51}= \frac{1}{13} * \frac{1}{17} = \frac{1}{221}\)
Answer: E
General Discussion
03 Nov 2015, 10:37
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Bunnel,
One thing, I didn't get it here, is
Why P= p1x p2?
why it can't be P = P1+P2.
I am missing some fundamentals thing. Please explain to clear my concept.
One thing, I didn't get it here, is
Why P= p1x p2?
why it can't be P = P1+P2.
I am missing some fundamentals thing. Please explain to clear my concept.
