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A box contains 4 red balls and 6 black balls. Three balls are selected randomly from the box one after another, without replacement. The probability that the selected set contains one red ball and two black balls is
- a)1/20
- b)1/12
- c)3/10
- d)1/2
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
A box contains 4 red balls and 6 black balls. Three balls are selected...
Solution:
Given, a box contains 4 red balls and 6 black balls.
Total balls = 4 + 6 = 10
Three balls are selected randomly from the box one after another, without replacement.
To find: Probability of selecting one red ball and two black balls.
Let A be the event of selecting one red ball and two black balls.
A can happen in the following ways:
1. RBB
2. BRB
3. BBR
Number of ways of selecting one red ball and two black balls = Number of ways of selecting 1 red ball from 4 red balls x Number of ways of selecting 2 black balls from 6 black balls
= 4C1 x 6C2
= 4 x 15
= 60
Total number of ways of selecting three balls from 10 balls = Number of ways of selecting 3 balls from 10 balls
= 10C3
= (10 x 9 x 8)/(3 x 2 x 1)
= 120
Therefore, the required probability is given by:
P(A) = Number of ways of selecting one red ball and two black balls/Total number of ways of selecting three balls from 10 balls
= 60/120
= 1/2
Hence, the correct option is D.
Given, a box contains 4 red balls and 6 black balls.
Total balls = 4 + 6 = 10
Three balls are selected randomly from the box one after another, without replacement.
To find: Probability of selecting one red ball and two black balls.
Let A be the event of selecting one red ball and two black balls.
A can happen in the following ways:
1. RBB
2. BRB
3. BBR
Number of ways of selecting one red ball and two black balls = Number of ways of selecting 1 red ball from 4 red balls x Number of ways of selecting 2 black balls from 6 black balls
= 4C1 x 6C2
= 4 x 15
= 60
Total number of ways of selecting three balls from 10 balls = Number of ways of selecting 3 balls from 10 balls
= 10C3
= (10 x 9 x 8)/(3 x 2 x 1)
= 120
Therefore, the required probability is given by:
P(A) = Number of ways of selecting one red ball and two black balls/Total number of ways of selecting three balls from 10 balls
= 60/120
= 1/2
Hence, the correct option is D.
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A box contains 4 red balls and 6 black balls. Three balls are selected randomly from the box oneafter another, without replacement. The probability that the selected set contains one red ball andtwo black balls isa)1/20b)1/12c)3/10d)1/2Correct answer is option 'D'. Can you explain this answer?
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