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# Two hard candies are to be selected at random and without replacement

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Intern
Joined: 16 Aug 2014
Posts: 49
Two hard candies are to be selected at random and without replacement  [#permalink]

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15 Oct 2018, 04:57
1
00:00

Difficulty:

5% (low)

Question Stats:

91% (01:04) correct 9% (02:19) wrong based on 44 sessions

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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

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Director
Joined: 19 Oct 2013
Posts: 519
Location: Kuwait
GPA: 3.2
WE: Engineering (Real Estate)
Re: Two hard candies are to be selected at random and without replacement  [#permalink]

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15 Oct 2018, 05:13
mayursurya wrote:
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

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 = $$\frac{6}{20} * \frac{5}{19}$$ = $$\frac{6}{4} * \frac{1}{19}$$ = $$\frac{3}{2} * \frac{1}{19}$$ = $$\frac{3}{38}$$

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Re: Two hard candies are to be selected at random and without replacement  [#permalink]

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15 Oct 2018, 05:19
mayursurya wrote:
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

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

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Joined: 24 Dec 2017
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Re: Two hard candies are to be selected at random and without replacement  [#permalink]

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15 Oct 2018, 08:39
Total candies = 20
Green = 5
Red = 6
Yellow = 9

Probability = Favourable/Total

=>Probability of selecting the first red candy = 6/20
=>Probability of selecting the second red candy = 5/19
=>6/20 x 5/19
=>30/380
=>3/38

Re: Two hard candies are to be selected at random and without replacement   [#permalink] 15 Oct 2018, 08:39
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