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16 Sep 2014, 00:13
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A box contains 11 hats, out of which 6 are red hats and 5 are green hats. If two hats are selected randomly without replacement, what is the probability that at least one green hat will be selected?
A. \(\frac{10}{11}\)
B. \(\frac{8}{11}\)
C. \(\frac{7}{12}\)
D. \(\frac{5}{13}\)
E. \(\frac{2}{7}\)
A. \(\frac{10}{11}\)
B. \(\frac{8}{11}\)
C. \(\frac{7}{12}\)
D. \(\frac{5}{13}\)
E. \(\frac{2}{7}\)
16 Sep 2014, 00:13
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Official Solution:
A box contains 11 hats, out of which 6 are red hats and 5 are green hats. If two hats are selected randomly without replacement, what is the probability that at least one green hat will be selected?
A. \(\frac{10}{11}\)
B. \(\frac{8}{11}\)
C. \(\frac{7}{12}\)
D. \(\frac{5}{13}\)
E. \(\frac{2}{7}\)
It's simpler to calculate the probability of the complementary event and subtract it from 1. The complementary event would be selecting no green hats, which is the same as selecting 2 red hats. Thus: \(P(\text{at least 1 green hat}) = 1 - P(\text{2 red hats}) = 1 - \frac{6}{11} \cdot \frac{5}{10} = \frac{8}{11}\).
Answer: B
A box contains 11 hats, out of which 6 are red hats and 5 are green hats. If two hats are selected randomly without replacement, what is the probability that at least one green hat will be selected?
A. \(\frac{10}{11}\)
B. \(\frac{8}{11}\)
C. \(\frac{7}{12}\)
D. \(\frac{5}{13}\)
E. \(\frac{2}{7}\)
It's simpler to calculate the probability of the complementary event and subtract it from 1. The complementary event would be selecting no green hats, which is the same as selecting 2 red hats. Thus: \(P(\text{at least 1 green hat}) = 1 - P(\text{2 red hats}) = 1 - \frac{6}{11} \cdot \frac{5}{10} = \frac{8}{11}\).
Answer: B
26 Jun 2017, 09:17
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Total Cases : 11C2 = 55
Prob. of Picking up One Green Hat = 5c1 * 6c1 = 30 (One from Green Hat Box and Other from Red Hat Box)
Prob. of Picking up Two Green Hat = 5c2 = 10 (Both from Green Hat Bucket)
P(x) = (30 + 10)/55
= 40/55 = 8/11
Prob. of Picking up One Green Hat = 5c1 * 6c1 = 30 (One from Green Hat Box and Other from Red Hat Box)
Prob. of Picking up Two Green Hat = 5c2 = 10 (Both from Green Hat Bucket)
P(x) = (30 + 10)/55
= 40/55 = 8/11
