Conditional Probability

[hollow bobcat]

could anyone help me understand question (b)

[ember acorn]

We have:
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
So, find these probabilities (one you've already found) and add them.

[hollow bobcat]

is it 0.975

[ember acorn]

Hm, no, I'm getting a different answer.

[hollow bobcat]

(0.15)^2 (0.85)^9 would be the probability that there is 2 left footed on the team?

[ember acorn]

No.
Suppose X is the random variable equal to the number of left-handed people on the team. Then:
P(X = k) = C(n, k) p^k (1 - p)^(n - k)
Here C(n, k) = n!/(k! (n - k)!) is the binomial coefficient. So, you forgot about C(n, k).

[hollow bobcat]

So its 11 total players / 2 who are left footed?

(0.15)^2 (0.85)^9 then this which is 9 right footed and 2 left foot

[ember acorn]

You re still forgetting C(n, k).

In our case X ~ Bin(n, p) with n = 11, p = 0.15 = 3/20.
P(X = 0) = C(n, 0) p^0 (1 - p)^(n - 0) = (17/20)^11
P(X = 1) = C(n, 1) p^1 (1 - p)^(n - 1) = 11(3/20)(17/20)^10
P(X = 2) = C(n, 2) p^2 (1 - p)^(n - 2) = 55(3/20)^2 (17/20)^9

[hollow bobcat]

Oh I see

(11/2)(0.15)^2(0.85)9 + 0.325 + (0.85)^11?

@ember acorn

[ember acorn]

Well, better to just enter the complete expressions in the calculator to avoid errors.

[hollow bobcat]

I feel like if I express it ill find it harder to remember

[ember acorn]

What do you mean? There's just one formula.

[hollow bobcat]

I mean I find it hard to remember the formulas

Theres about 7 for probability so its a bit hard

[ember acorn]

Well, there will be more.

There are a lot of distributions, even common ones.

[hollow bobcat]

Is the one I just used now Bayes?

[ember acorn]

No, it's the PMF of binomial distribution.

So, this isn't even conditional probability. Not sure why you called it that.

[hollow bobcat]

oh sorry

[hollow bobcat]

thank you anyways