Question

Assume that Tom attends randomly with probability 0.55 and that each decision is independent of previous attendance. What is the probability that he attends at least 7 of 10 classes given that he attends at least 2 but not all 10 classes?

Answer 1

I got `~~0.26350/0.99296~~0.26537~~26.54%`

Explanation:

Let's first work on the binomial probability of Tom attending his classes.

We'll be using this relation:

`sum_(k=0)^(n)C_(n,k)(p)^k(1-p)^(n-k)=1`

Looking at the full range of probabilities for Tom attending class, we'll have `n=10`, `k=` the number of classes he actually attends, and `p=0.55`. And so the full relation is:

`sum_(k=0)^(10)C_(10,k)(0.55)^k(0.45)^(n-k)=1`

So now let's get more into the specifics. We're looking for the probability that he attends class, but it's a conditional probability - we're told he attends at least 2 but not all 10 - which means the probability of his attending "all his classes" isn't 1 but something less than that - we need to subtract out the classes we know he won't attend and set that as the denominator:

`1-(C_(10,0)(0.55)^0(0.45)^(10)+C_(10,1)(0.55)^1(0.45)^(9)+C_(10,10)(0.55)^10(0.45)^(0))~~1-0.00704~~0.99296`

Now the numerator. We're asked for the probability that he attend at least 7 of his classes, but we're also told he won't attend all 10, and so we'll sum up for `7<=k<=9`

`C_(10,7)(0.55)^7(0.45)^(3)+C_(10,8)(0.55)^8(0.45)^(2)+C_(10,9)(0.55)^9(0.45)^(1)~~0.26350`

And so the probability that he attends at least 7 classes, knowing he'll attend at least 2 but not 10 is:

`~~0.26350/0.99296~~0.26537~~26.54%`