trankimphuong wrote:
A box contains 10 red balls and 30 blue balls. Balls will be pulled out of the box and then put back. It will stop when 2 blue balls have been pulled out of the box. What is the probability of getting at least 1 red ball before stopping?
A. 1/2
B. 2/3
C. 9/6
D. 3/4
E. 7/6
Choices are clearly flawed. You can't have probability >1. Both C(9/6) and E(7/6) are >1..
So it should be 16 in denominator..
Easiest way is to find prob of both being BLUE, meaning NO red .
Probability of B one after another = \(\frac{30}{10+30}*\frac{30}{10+30}=\frac{3*3}{4*4}=\frac{9}{16}\)
This is the only way that Red is not picked up before the game stops. Otherwise the ways can be
1) B,R,R,B
2)R,R,B,B
3) R,R,R,R,R,B,R,B
And so on
So probability that atleast one red is picked = 1-(probability that NO red is picked)=1-9/16=7/16
E
_________________