Question

There are 8 boys and 7 girls in a class room. If the names of all those children are written on paper slips and 3 slips are drawn at random from them, then the probability of getting the names of one boy and two girls or one girl and two boys is

• A. \(\frac{1}{5}\)
• B. \(\frac{3}{4}\)
• C. \(\frac{4}{5}\)
• D. \(\frac{1}{4}\)

Hint:

When solving probability problems involving combinations (choosing items without regard to order), remember to: Calculate the total number of possible outcomes using \(C(n,r)\). Calculate the number of favorable outcomes for each specific case described. If the cases are mutually exclusive (cannot happen at the same time, like "1 boy and 2 girls" versus "1 girl and 2 boys" when drawing 3 slips), add the number of ways for each case to find the total favorable outcomes. Divide the total favorable outcomes by the total possible outcomes to get the probability.

Answer 1

The Correct Option is C

Step 1: Calculate the total number of possible outcomes.
Total number of boys = 8
Total number of girls = 7
Total number of children = 8 + 7 = 15.
We need to draw 3 slips from these 15 children. The total number of ways to do this is given by the combination formula \(C(n, r) = \frac{n!}{r!(n-r)!}\): \[ \text{Total outcomes} = C(15, 3) = \frac{15!}{3!(15-3)!} = \frac{15!}{3!12!} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 5 \times 7 \times 13 = 455. \] Step 2: Calculate the number of favorable outcomes for "one boy and two girls".
Number of ways to choose 1 boy from 8 boys = \(C(8, 1) = \frac{8!}{1!(8-1)!} = 8\).
Number of ways to choose 2 girls from 7 girls = \(C(7, 2) = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21\).
The number of ways to get one boy and two girls is the product of these combinations: \[ \text{Ways (1 boy, 2 girls)} = C(8, 1) \times C(7, 2) = 8 \times 21 = 168. \] Step 3: Calculate the number of favorable outcomes for "one girl and two boys".
Number of ways to choose 1 girl from 7 girls = \(C(7, 1) = \frac{7!}{1!(7-1)!} = 7\).
Number of ways to choose 2 boys from 8 boys = \(C(8, 2) = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28\).
The number of ways to get one girl and two boys is the product of these combinations: \[ \text{Ways (1 girl, 2 boys)} = C(7, 1) \times C(8, 2) = 7 \times 28 = 196. \] Step 4: Calculate the total number of favorable outcomes.
The problem asks for the probability of "one boy and two girls" OR "one girl and two boys". Since these two events are mutually exclusive, the total number of favorable outcomes is the sum of the ways calculated in Step 2 and Step 3: \[ \text{Total favorable outcomes} = 168 + 196 = 364. \] Step 5: Calculate the probability.
The probability is the ratio of the total favorable outcomes to the total possible outcomes:
\[ \text{Probability} = \frac{\text{Total favorable outcomes}}{\text{Total outcomes}} = \frac{364}{455}. \] Simplify the fraction by dividing both numerator and denominator by their greatest common divisor.
Both are divisible by 7:
\(364 \div 7 = 52\)
\(455 \div 7 = 65\)
So the fraction becomes \(\frac{52}{65}\).
Both are divisible by 13:
\(52 \div 13 = 4\)
\(65 \div 13 = 5\)
So, the probability is \(\frac{4}{5}\).