ps: jars (probability)

Question

A small jar contains 6 balls, 3 of which are red. Another jar contains 8 balls, 6 of which are red. Lisa randomly takes two balls from the first jar and places it them the second, and Ray randomly takes two balls from the second and places them into the first. Finally, Dean randomly takes two balls from the first jar. Which of the following best approximates the probability that all six balls that were removed from the jars were red?

1/50 1/40 1/30 1/20 1/15

1/50 1/40 1/30 1/20 1/15

maratikus · Answer

probability of picking 2 red balls from the first jar 3C2/6C2
now we have 8 red balls and 2 other balls in the second jar -> probability of picking 2 red balls from the second jar is 8C2/10C2
now we have 5 red balls and 1 other ball in the first jar -> 5C2/6C2

probability that all 6 balls were red (3C2/6C2)*(8C2/10C2)*(5C2/6C2) = (3*2/6*5)*(8*7/10*9)*(5*4/6*5) = (8*7*4/5*10*9*6) = approx = 0.08 -> E

walker · Answer

B

p= * * =\frac{28}{25*45}\approx\frac{1}{40}

GMATBLACKBELT · Answer

3/6*2/5 
*
8/10*7/9
*
3/6*2/5

1/4*4/25*4/5*7/9 --> 28/1125 ~ 30/1000--> 3/100 realize this is an overestimate. so its btwen 1/50 and 1/40

1/50 is too small thats pretty much 20/1000. SO 1/40 seems to be it.

B

maratikus · Answer

you are right, i miscalculated the third event ...

pmenon · Answer

i did it like walker and gmatblackbelt, just calculate the individual probabilities.

but i wasnt sure whether we assume that we put 2 more reds into the second jar, i.e. will probability be 8/10. or 6/10 ....

walker · Answer

We have dependent event. Fortunately, the all-red condition makes the problem easier.
We know that Lisa namely add 2  red balls  into second jar. Therefore, we have 10 balls, 8 of which are red.

ps: jars (probability)

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