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04 Jan 2021, 22:15
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Difficulty:
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92% (01:03) correct
8%
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based on 85
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A bag contains 2 red beads, 2 blue beads, and 2 green beads. Sara randomly draws a bead from the bag, without replacement and then Victor randomly draws a bead from the bag. What is the probability that Sara will draw a red marble and Victor will draw a blue marble?
A. 1/3
B. 1/5
C. 2/13
D. 2/15
E. 1/10
A. 1/3
B. 1/5
C. 2/13
D. 2/15
E. 1/10
04 Jan 2021, 22:42
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[A bag contains 2 red beads, 2 blue beads, and 2 green beads. Sara randomly draws a bead from the bag, without replacement and then Victor randomly draws a bead from the bag. What is the probability that Sara will draw a red marble and Victor will draw a blue marble?]
The bag contains 6 beads in total.
Sarah choosing a red bead is 2 out of 6
As Sarah did not put it back, there are left 5 beads in the bag.
Victor choosing a blue bead is 2 out of 5
both happening simultaneously is (2/6)x(2/5) = 4/30 = 2/15
IMHO
The bag contains 6 beads in total.
Sarah choosing a red bead is 2 out of 6
As Sarah did not put it back, there are left 5 beads in the bag.
Victor choosing a blue bead is 2 out of 5
both happening simultaneously is (2/6)x(2/5) = 4/30 = 2/15
IMHO
XSatishX
Joined: 05 Jul 2020
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04 Jan 2021, 23:29
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We have 2 red, 2 blue and 2 green beads.
Probability of first red marble = 2/6 = 1/3
Probability of second blue marble = 2/5
Probability of A and B = P(A) * P(B)
P = 1/3*2/5 = 2/15
Answer D
Probability of first red marble = 2/6 = 1/3
Probability of second blue marble = 2/5
Probability of A and B = P(A) * P(B)
P = 1/3*2/5 = 2/15
Answer D
