Question

A baseball player has a batting average of 0.280.
(i) at least 3 hits in her next 5 times at bat
(ii) at least 3 hits in her next 10 times at bat
(iii) at least 6 hits in her next 10 times at bat
(b) What is the player’s expected number of hits in her next 10 times at bat?
I can only figure out the first one. I can't figure out the rest and I don't know what I'm, doing wrong. I tried doing the compliments. I can't figure it out. Can someone help?

Answer 1

it wants the probability

Answer 2

b) 10 * 0.28

Answer 3

Here's a diagram showing the outcomes for 3 at bats.

Answer 4

Notice that there is 1 chain that leads to 3 hits, 3 chains that lead to 2 hits, 3 chains that lead to 1 hit, and 1 chain that leads to 0 hits. Does that sequence of numbers (1 3 3 1) ring any bells?

Answer 5

Here's a hint: if we extend that diagram out to 4 at bats, the sequence would be

1 4 6 4 1

Answer 6

And if we went to 5 at bats, we'd have

1 5 10 10 5 1

Answer 7

These are just the binomial coefficients. Referring back the diagram I made, we can see that the P(3 hits in 3 at bats) = 0.28 * 0.28 * 0.28 = 0.022.

What if we wanted to find the probability of 10 hits in 10 at bats? Well, that's 0.28^10 = 2.96^10^-6 — call it 3 in a million :-) The important point is we didn't have to make the chart 10 levels deep to figure that out. We just multiplied the probability of a hit with itself enough times.

Answer 8

Now, if we have looser standards, and we only need n hits in m at bats, we just have to figure out which chains result in at least n hits.

Let's look at your first one, 3 hits in 5 at bats.

we can get 5 hits in 5 at bats, with P = 0.28^5 = 0.0017
We can get 4 hits in 5 at bats, with P = 0.28^4*0.72 = 0.0044
We can get 3 hits in 5 at bats, with P = 0.28^3*0.72^2 = 0.0114

Now, we have to adjust those probabilities, because we didn't account for the fact that it is possible to get 3 or 4 hits in 5 at bats in a number of different ways: we just computed the probability of getting 3 hits in a row followed by 2 non-hits, or 4 hits in a row followed by 1 non-hit. We can do this adjustment by multiplying each of those probabilities by the corresponding binomial coefficient. For 5 at bats, the series is

1 5 10 10 5 1
corresponding to
5 hits
4 hits
3 hits
2 hits
1 hit
0 hits

so our P(5 hits) is multiplied by 1, giving us 0.0017
P(4 hits) is multiplied by 5, giving us 0.022
P(3 hits) is multiplied by 10, giving us 0.114

we add the three together to get the total probability of getting 3 hits in 5 at bats with a batting average of 0.28, and the result is P = 0.1377

If we added the rest of the probabilities:
P(2 hits) = 10*0.28^2*0.72^3 = 0.293
P(1 hit) = 5*0.28^1*0.72^4 = 0.376
P(0 hits) = 1*0.72^5 = 0.193

add those to our P(3,4,5 hits) and it should come out to 1.0, barring any rounding issues.

Answer 9

Do you see how to tackle the remaining problems now?

Answer 10

Thank you so much! That makes so much more sense now. I swear you explained it better than my math teacher!

Answer 11

Can I give you a billion metals!

Answer 12

What's a testimonial? Sorry, even though I've had this account for a while, I never had time to actually get to explore how it works.

Answer 13

oh okay. So what would I write exactly? I feel like a child right now