Sajjad1994 wrote:
Balls with consecutive integers from 0 to x are placed in a bowl. Peter draws a ball from the bowl and records the number. Without replacing the first ball, he draws a second ball from the bowl. If the probability that Peter does NOT draw an odd numbered ball on either draw is 12/42, what integer is on the highest numbered ball?
A. 3
B. 4
C. 5
D. 6
E. 7
When we say that the probability that Both are NOT odd numbers means that both are even numbers. So probability of drawing 2 even numbered balls without replacement = \(\frac{12}{42}\)
If x is the largest integer, then the total number of balls is x + 1. For e.g if we have x = 3, then the balls are numbered from 0 till 3 i.e 0,1,2 and 3.
Here is where we can get our first clue. The probability is given as \(\frac{12}{42}\) and since we are picking without replacement, the total outcomes in the denominator will be (x + 1) * (x).
42 can be written as the product of 2 consecutive integers as 7 * 6. Therefore the answer should be 6.
We can check. If x = 6, then the balls are numbered as 0, 1, 2, 3, 4, 5 and 6. These have 4 even numbers.
Probability of picking 2 even numbers without replacement = \(\frac{4}{7} * \frac{3}{6} = \frac{12}{42}\). Therefore x = 6 is correct.
Option DArun Kumar
_________________
Crackverbal Prep Team
www.crackverbal.com