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04 Jan 2021, 05:24
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E
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Difficulty:
25%
(medium)
Question Stats:
78% (01:33) correct
22%
(02:23)
wrong
based on 54
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Not Attempted Yet
Seven beads are in a bag: three blue, two red, and two green. If three beads are randomly drawn from the bag, what is the probability that they are not all blue?
A. 8/13
B. 5/7
C. 6/7
D. 23/24
E. 34/35
A. 8/13
B. 5/7
C. 6/7
D. 23/24
E. 34/35
rocky620
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04 Jan 2021, 05:39
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Total Cases = 7c3 = 35
P of selecting max 2 blue marbles (since not all can be blue)= 1 - Probability of selecting all blue
P (all blue) = 3/7 * 2/6 * 1/5 = 1/35
P of selecting max 2 blue marbles = 1 - 1/35 = 34/35
IMO Option E
P of selecting max 2 blue marbles (since not all can be blue)= 1 - Probability of selecting all blue
P (all blue) = 3/7 * 2/6 * 1/5 = 1/35
P of selecting max 2 blue marbles = 1 - 1/35 = 34/35
IMO Option E
04 Jan 2021, 05:47
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The answer is E
"Not all blue" means none of the three balls you draw are blue.
The probability of drawing all blue balls from the bag, say P(B)
Probability of none of the balls being blue will then be 1-P(B)
P(B)=3/7*2/6*1/5 = 1/35
So, the probability of none of the balls being blue = 1-1/35 = 34/35 (Answer E)
"Not all blue" means none of the three balls you draw are blue.
The probability of drawing all blue balls from the bag, say P(B)
Probability of none of the balls being blue will then be 1-P(B)
P(B)=3/7*2/6*1/5 = 1/35
So, the probability of none of the balls being blue = 1-1/35 = 34/35 (Answer E)
