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Probability that Jean drew a pink ball

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Probability that Jean drew a pink ball

by Vemuri » Sun Mar 22, 2009 7:32 am
Jean drew a gumball at random from a jar of pink and blue gumballs. Since the gumball she selected was blue and she wanted a pink one, she replaced it and drew another. The second gumball also happened to be blue and she replaced it as well. If the probability of her drawing the two blue gumballs was 9/49, what is the probability that the next one she draws will be pink?

A. 1/49
B. 4/7
C. 3/7
D. 16/49
E. 40/49
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by DanaJ » Sun Mar 22, 2009 8:14 am
I'm gonna give it my best shot, but I'm not that good with probabilities, so....
Say you have:
x = number of blue gumballs
y = number of pink gumballs
t = total number of gumballs

The probability of the first gumball being blue is x/t. Since she places it in the jar, then the probability of the second gumball being blue is again x/t (since we're basically in the same spot, after putting the gumball back).

This means that the probability of her taking out the two gumballs will be x/t * x/t = (x^2)/(t^2) = 9/49, making x/t = 3/7.

Now, the probability of the third gumball that she draws will be 1 - 3/7 = 4/7 (IMHO, again). This happens because she put the second blue gumball back and she's again back to square one. In this third attempt, again we have that the probability of taking out a blue ball is x/t = 3/7, so the probability that a pink gumball will be selected is 1 - 3/7.
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by lilu » Sun Mar 22, 2009 12:49 pm
1st AND 2nd attempt: Probability of picking B*Probability of picking B=9/49
So, probability of picking blue is 3/7; this means that there are 3 blue balls and 7 balls in total--> 4 pink balls
So prob of picking pink ball is 4/7
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