Question

Carl has a piece of toast that has jelly on one side, and he dropped it twice. Both times, it landed with the jelly side up. If he drops it two more times, what is the probability that it will have landed jelly side up a total of three times?

Answer 1

`.25`

Explanation:

The amount of ways the toast can land jelly side up 3 out of 4 times using the binomial distribution we can look at this as the total number of success from 4 trials with probability `p=1/2` thus

`("_k^n)p^k(1-p)^(n-k)`

`("_3^4)(1/2)^3(1-1/2)^(4-3)`

`4(1/2)^4 = 1/4 = .25`

You can also go through the combinations for this and see that
the total number of combinations of front to back is `2^4` which is the denominator and then the amount of ways you can get 3 out of 4 is `("_3^4)` is the numerator

`(("_3^4))/2^4 =.25`

Answer 2

I say 1/2.

Explanation:

I see that this probability answer is about 1 year old, and no one has challenged it. But ... here goes.

The way the question is worded, the first 2 chances have already happened and it was jelly side up both times. The question asks about getting jelly side up once out of the next 2 chances.

Notation: 0 is jelly down and 1 is jelly up.
The possibilities of the next 2 chances are:
00
01
10
11

Of the 4 equal possibilities, both 10 and 01 would give a total of 3 times that it was jelly side up.

So I say 1/2. Comments?
Steve

P.S. I should admit that I have very little qualifications in probability. But I think the issue here is interpretation of the question.