Mr. Hauseman has 17 students in his class, three of whom are freshmen, and the rest are from other classes. He is going to draw two students randomly to be partners.
He calculates the probability of drawing a freshman and then a junior to be
9
136
. How many juniors must be in the class?
A) 3
B) 6
C) 7
D) 8
Answers
Answer:
6 852369514452878621
There must be 6 juniors in Mr. Hauseman's class. Therefore, option (B) is correct.
• Given,
Number of students in Mr. Hauseman's class = 17
Number of freshmen in the class = 3
Probability of drawing a freshman and then a junior (P) = 9 / 136
• Now, the probability of drawing a freshman first = 3 / 17
Number of students left to be selected = 17 - 1 = 16
• Let the number of juniors in the class be x.
Then, the probability of selecting a junior next = x / 16
• Therefore, probability of selecting a freshman, followed by a junior (P) = Probability of selecting a freshman × Probability of selecting a junior
=> P = (3 / 17) × (x / 16)
=> 9 / 136 (given) = (3 / 17) × (x / 16)
=> 9 / 136 = (3 × x) / (17 × 16)
=> 9 / 136 = 3x / (17 × 16)
On making x the subject of formula by cross-multiplication, we get,
(9 × 17 × 16) / (136 × 3) = x
=> x = (9 × 17 × 16) / (136 × 3)
=> x = 3 × 2 (After cancelling out all the common factors)
=> x = 6
∴ The number of juniors in Mr. Hauseman's class is 6 (option B).