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A bag contains 3 gold coins and 7 silver coins. The bag cont
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05 Oct 2018, 23:39
05 Oct 2018, 23:39
Expert Reply
00:00
Question Stats:
58% (01:57) correct
41% (01:41) wrong based on 34 sessions
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A bag contains 3 gold coins and 7 silver coins. The bag contains nothing else. If 3 coins are selected at random and without replacement from the bag, what is the probability that exactly 2 of the coins are silver?
A. \(\frac{7}{40}\)
B. \(\frac{7}{24}\)
C. \(\frac{3}{10}\)
D. \(\frac{21}{40}\)
E. \(\frac{2}{3}\)
A. \(\frac{7}{40}\)
B. \(\frac{7}{24}\)
C. \(\frac{3}{10}\)
D. \(\frac{21}{40}\)
E. \(\frac{2}{3}\)
Re: A bag contains 3 gold coins and 7 silver coins. The bag cont
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05 Oct 2018, 23:50
05 Oct 2018, 23:50
Carcass wrote:
A bag contains 3 gold coins and 7 silver coins. The bag contains nothing else. If 3 coins are selected at random and without replacement from the bag, what is the probability that exactly 2 of the coins are silver?
A. \(\frac{7}{40}\)
B. \(\frac{7}{24}\)
C. \(\frac{3}{10}\)
D. \(\frac{21}{40}\)
E. \(\frac{2}{3}\)
A. \(\frac{7}{40}\)
B. \(\frac{7}{24}\)
C. \(\frac{3}{10}\)
D. \(\frac{21}{40}\)
E. \(\frac{2}{3}\)
the probability of picking exactly 2 silver and 1 gold
SSG, SGS, GSS , i.e. they can be arranged in 3 ways
Now,
Probability of picking exactly 2 silver and 1 gold = \(3 * \frac{3}{10} * \frac{7}{9} * \frac{6}{8} = \frac{21}{40}\)
Re: A bag contains 3 gold coins and 7 silver coins. The bag cont
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07 Oct 2018, 06:00
07 Oct 2018, 06:00
Expert Reply
Do not get the first 3
??
??
