Question
Download Solution PDFComprehension
Answer these questions (Qs. 77 - 79) based on the following information.
There are 4 red balls, 4 green balls and 6 blue balls, in a box
If three balls are drawn randomly, what is the probability that one of them is green and the other two are blue?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCalculation:
Total Number of Balls and Possible Outcomes
Total balls: 4 (red) + 4 (green) + 6 (blue) = 14 balls
We're drawing 3 balls, so the total number of possible combinations is given by the combination formula:
¹⁴C₃ = (14!)/(3! × 11!) = (14 × 13 × 12)/(3 × 2 × 1) = 364
Favorable Outcomes
We want 1 green ball and 2 blue balls.
Number of ways to choose 1 green ball from 4: ⁴C₁ = 4
Number of ways to choose 2 blue balls from 6: ⁶C₂ = (6 × 5)/(2 × 1) = 15
To get the total number of favorable combinations, we multiply these: 4 × 15 = 60
Calculate the Probability
Probability = (Favorable Outcomes) / (Total Possible Outcomes)
Probability = 60 / 364
Simplify the Fraction
Divide both numerator and denominator by 4: 15/91
Answer:
The probability of drawing one green ball and two blue balls is 15/91.
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