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A box contains 12 balls, 7 of them are red and 5 of them are green. If
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04 Dec 2023, 00:58
04 Dec 2023, 00:58
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Question Stats:
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42% (02:28) wrong based on 7 sessions
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A box contains 12 balls, 7 of them are red and 5 of them are green. If 3 balls are to be selected at random from the box, what is the probability that 2 balls will be red and 1 ball will be green?
A. 7/44
B. 7/22
C. 51/100
D. 21/44
E. 7/9
A. 7/44
B. 7/22
C. 51/100
D. 21/44
E. 7/9
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A box contains 12 balls, 7 of them are red and 5 of them are green. If
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04 Dec 2023, 19:46
04 Dec 2023, 19:46
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There are 3 different ways you could pick the balls: 1) Red, Red, Green vs. 2) Red, Green, Red vs. 3) Green, Red, Red.
Let's pick the first scenario. There are 7 out 12 Reds. Thus, the probability of picking a Red is 7/12. The probability of picking another Red is 6/11, since 6 Reds remain and there are now 11 total balls. The probability of picking a Green is 5/10.
\(\frac{7}{12} * \frac{6}{11} * \frac{5}{10}\)
\(= \frac{210}{1320}\)
\(= \frac{7}{44}\)
However, since there are 3 different ways, multiply by 3:
\(= \frac{7}{44}*3\)
\(= \frac{21}{44}\)
Let's pick the first scenario. There are 7 out 12 Reds. Thus, the probability of picking a Red is 7/12. The probability of picking another Red is 6/11, since 6 Reds remain and there are now 11 total balls. The probability of picking a Green is 5/10.
\(\frac{7}{12} * \frac{6}{11} * \frac{5}{10}\)
\(= \frac{210}{1320}\)
\(= \frac{7}{44}\)
However, since there are 3 different ways, multiply by 3:
\(= \frac{7}{44}*3\)
\(= \frac{21}{44}\)
gmatclubot
A box contains 12 balls, 7 of them are red and 5 of them are green. If [#permalink]
04 Dec 2023, 19:46
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