PS - Good probablity question

Question

Tricky question, thought I'd share with the group. Took me 3 mins to solve

A fair, six-sided die, with sides numbered one through six, is rolled 5 times. What is the probability that on exactly 3 rolls the number of dots showing is greater than 4? A) 1/243 B) 10/243 C) 1/27 D) 40/243 E) 80/243 edit : I mistyped, it should say "What is the probability that on exactly..."

jallenmorris · Answer

Please rephrase the question. It's too confusing to be able to answer properly.

Oski · Answer

I think answer is (D)

In order to have more than 4 on exactly 3 rolls, you have to get 5 or 6 on 3 rolls (chance this happens for 1 roll: 2/6) and 1,2,3 or 4 on the 2 others (chance this happens for 1 roll: 4/6).

Once you chose which of the 3 rolls will bear the 5 or 6, you have (2/6)^3 * (4/6)^2 chances it happens

This is 4/243

And there are 10 choices of the 3 rolls on 5 (unordered choices).

Therefore answer is 40/243.

walker · Answer

Agree with Oski

p=C^5_3*(\frac{2}{6})^3*(\frac{4}{6})^2=\frac{5!*2^2}{3!*2!*3^5}=\frac{40}{243}

xALIx · Answer

For this question, the answer is 40/243, the key is including the combinations so 4/243*10

xALIx · Answer

I'll state the question differently, for similar practice:

A fair, six-sided die, with sides numbered one through six, is rolled 5 times. What is the probability that on exactly 3 rolls the number of dots showing is  no greater  than 4?
A) 1/243
B) 10/243
C) 1/27
D) 40/243
E) 80/243

btw, the answer is different from the one above

Oski · Answer

Same reasoning gives :
 p=C^5_3*(\frac{4}{6})^3*(\frac{2}{6})^2=\frac{5!}{3!*2!}*\frac{2^3}{3^5}=\frac{80}{243}

PS - Good probablity question

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