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15 Oct 2018, 04:57
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E
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Difficulty:
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Question Stats:
90% (01:06) correct
10%
(02:09)
wrong
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Two hard candies are to be selected at random and without replacement from a bag containing 5 green, 6 red, and 9 yellow candies. What is the probability that both candies selected will be red?
A. 3/10
B. 9/49
C. 15/98
D. 9/100
E. 3/38
A. 3/10
B. 9/49
C. 15/98
D. 9/100
E. 3/38
15 Oct 2018, 05:13
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mayursurya
Two hard candies are to be selected at random and without replacement from a bag containing 5 green, 6 red, and 9 yellow candies. What is the probability that both candies selected will be red?
A. 3/10
B. 9/49
C. 15/98
D. 9/100
E. 3/38
A. 3/10
B. 9/49
C. 15/98
D. 9/100
E. 3/38
Total possible selection = 5+6+9 = 20
Total possible red = 6
The question asks for 2 red and it says without replacement so it will be in the form of \(\frac{R}{Total} * \frac{(R -1)}{Total}\)
Probability of selecting 2 red = \(\frac{6}{20} * \frac{5}{19}\) = \(\frac{6}{4} * \frac{1}{19}\) = \(\frac{3}{2} * \frac{1}{19}\) = \(\frac{3}{38}\)
Answer choice E
15 Oct 2018, 05:19
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mayursurya
Two hard candies are to be selected at random and without replacement from a bag containing 5 green, 6 red, and 9 yellow candies. What is the probability that both candies selected will be red?
A. 3/10
B. 9/49
C. 15/98
D. 9/100
E. 3/38
A. 3/10
B. 9/49
C. 15/98
D. 9/100
E. 3/38
Total candies = 5+6+9 = 20
Red candies = 6
Favourable Outcomes = Total ways of selecting two red candies (out of 6) = 6C2 = 15
Total Outcomes = Total ways of selecting two candies (out of total 20 candies) = 20C2 = 190
Probability = favourable Outcomes / Total Outcomes = 15/190 = 3/38
Answer: Option E
