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Help with Binomial Probability (1 Viewer)
- Thread starter JaxsenW
- Start date
Edit: only need a solution for 19c)Hey guys, just wondering if someone could do a solution for this, I can't seem to get there, and neither can anyone I have asked...
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Others were easy
Thanks
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hahaha literally went through the question with my teacher today
First of all, there's a total of 4 hits and the 3 aces can occur in any order, therefore that occurs 4C3 times which is 4.
Just looking at one case (ace, ace, ace, in), there's a 1/15 chance of hitting an ace, so for the three aces thats 1/15 x 1/15 x 1/15.
But with the other serve that just lands in, we can't just multiply it by 8/10 because we can't ensure that it's going to be an ace, so you have to do 8/10-1/15 = 11/15
Therefore, you multiply 1/15 x 1/15 x 1/15 x 11/15 x 4C3 to get the answer which is 0.000869
Sorry if that was hard to follow. I have the working out written down but wasn;t able to send a photo
First of all, there's a total of 4 hits and the 3 aces can occur in any order, therefore that occurs 4C3 times which is 4.
Just looking at one case (ace, ace, ace, in), there's a 1/15 chance of hitting an ace, so for the three aces thats 1/15 x 1/15 x 1/15.
But with the other serve that just lands in, we can't just multiply it by 8/10 because we can't ensure that it's going to be an ace, so you have to do 8/10-1/15 = 11/15
Therefore, you multiply 1/15 x 1/15 x 1/15 x 11/15 x 4C3 to get the answer which is 0.000869
Sorry if that was hard to follow. I have the working out written down but wasn;t able to send a photo