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A basket contains 5 red 4 blue 3 green marbles. If three marbles picked up random, What is the probability that at least one is blue?
  • a)
    41/55
  • b)
    53/55
  • c)
    47/55
  • d)
    49/55
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A basket contains 5 red 4 blue 3 green marbles. If three marbles picke...
Total Marbles = 12
other than blue 8c3 / 12c3 = 14/55
probability = 1-14/55 = 41/55
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A basket contains 5 red 4 blue 3 green marbles. If three marbles picke...
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A basket contains 5 red 4 blue 3 green marbles. If three marbles picke...
Problem:
A basket contains 5 red marbles, 4 blue marbles, and 3 green marbles. If three marbles are picked at random, what is the probability that at least one of them is blue?

Solution:
To find the probability of at least one blue marble being picked, we need to consider all possible scenarios where at least one blue marble is included in the selection.

Total number of ways to select 3 marbles:
The total number of ways to select 3 marbles from a basket of 12 marbles is given by the combination formula:

C(12, 3) = 12! / (3! * (12-3)!) = 12! / (3! * 9!) = (12 * 11 * 10) / (3 * 2 * 1) = 220

Therefore, there are 220 possible ways to select 3 marbles from the basket.

Number of ways to select 3 marbles with at least one blue marble:
To find the number of ways to select 3 marbles with at least one blue marble, we need to consider two scenarios:
1. Selecting exactly one blue marble and two non-blue marbles.
2. Selecting exactly two blue marbles and one non-blue marble.

Scenario 1: Selecting exactly one blue marble and two non-blue marbles
Number of ways to select one blue marble: C(4, 1) = 4
Number of ways to select two non-blue marbles: C(8, 2) = 8! / (2! * (8-2)!) = (8 * 7) / (2 * 1) = 28
Therefore, the number of ways to select three marbles with exactly one blue marble is 4 * 28 = 112.

Scenario 2: Selecting exactly two blue marbles and one non-blue marble
Number of ways to select two blue marbles: C(4, 2) = 4! / (2! * (4-2)!) = (4 * 3) / (2 * 1) = 6
Number of ways to select one non-blue marble: C(8, 1) = 8
Therefore, the number of ways to select three marbles with exactly two blue marbles is 6 * 8 = 48.

Total number of ways to select 3 marbles with at least one blue marble:
The total number of ways to select three marbles with at least one blue marble is the sum of the scenarios calculated above:
112 + 48 = 160.

Probability of at least one blue marble being picked:
The probability is given by the number of favorable outcomes (selecting three marbles with at least one blue marble) divided by the total number of possible outcomes (selecting three marbles from the basket).

Probability = 160 / 220 = 8/11 = 41/55

Therefore, the correct answer is option 'A' (41/55).
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