Problem Solving

Molly needs to roll a 1 in a die and she has 3 tries. What is the probability that she will succeed?


can some body answer this with an explanation.

Regards
Naren

Hi All,

Thanks for passing by, found the answer. If anyone is interested below is how.

Probability to roll a 1 at the first try: 1/6
Probability to roll a 1 at the second try: 5/6 * 1/6 = 5/36
Probability to roll a 1 at the third try: 5/6 * 5/6 * 1/6 = 25/216
Since she can succeed with one OR two OR three tries, we sum it: 91/216

P(2nd roll) is the probability that you DON'T roll a 1 on the 1st roll, because the only way to get to the 2nd roll is to avoid a 1 on the 1st roll. this is why we need to multiply the 1/6 by 5/6.



Regards
Naren

Thanks for the reply Brent.

Regards
Naren

Just to illustrate why Brent's approach is so good, consider the positive approach, in which we list all the cases.

To get exactly one 1, we would need to roll 1, not-1, not-1 in some order. That gives us

1XX = (1/6) * (5/6) * (5/6) = 25/216

But this could happen in any of THREE orders: 1XX, X1X, XX1. So we have 3 * (25/216) = 75/216.

Now we consider exactly two 1s, which is basically the same idea:

11X = (1/6) * (1/6) * (5/6) = 5/216

This could happen in three orders as well (11X, 1X1, X11), so 3 * (5/216) = 15/216.

We also have exactly three ones, which is 111 = (1/6)*(1/6)*(1/6) = 1/216.

Adding these up, we have 75/216 + 15/216 + 1/216 = 91/216. Yuck!

So doing P(At least one) = 1 - P(No ones) is much more efficient.

Yeah agree matt this is so direct

In probability, we have a distribution called Geometric Distribution.
P(X = x) = p.(q)^(x - 1)
p = probability of success
q = probability of failure

Here we have,
P(1《 X 《 3) = P(X = 1) + P(X = 2) + P(X = 3)
= p + pq + pq^2 = (1/6) + (1/6)(5/6) + (1/6)(5/6)^2 = 91/216

Hi talaangoshtari,

how do we need to determine the usage of this neat little formula.

Regards
Naren

Hi NarendraSure,

Geometric distribution computes the probabability of the first success after k - 1 failures. In this problem there are 3 scenarios:

1. We roll a 1 in a die in the first try, so we have one success.
2. We roll a 1 in a die in the second try, so we have one failure and one success.
3. We roll a 1 in a die in the third try, so we have two failures and one success.

Brent's way to solve is simple calculation. Great.

But it is difficult to strike. But it is very good to solve such questions

P(A) = 1- P(Not A)