Question

Pete, a professional bowler, is unhappy with any game below 200. Over time, 80% of his games exceed this score. What is the probability that Pete exceeds 200 in at least 9 of his next 10 games?

Answer 1

I have:
P(9/10 games> 200) +P(10/10 grames> 200)
Or
1-P(Doesn't exceed any or doesn't exceed 1 game)

Answer 2

1st way is easiest on this one.

Answer 3

Do I use combinatorial

Answer 4

if you do the 1 - P thing you'll have to calulate
P(8/10 >200) P( 7/10>200), ... all the way down to P(0/10 >200)

Answer 5

at least 9 means nine or ten
easier than computing
none
one
two
three
four
five
six
seven
eight

Answer 6

use the binomial to compute this
but the only combination you need is ten choose 9 which is ten

Answer 7

Start with the easier one.
What's P(10/10 games are >200)?

Answer 8

.8^10?

Answer 9

yes

Answer 10

right.
and P(9/10 games are >200) = ?
Any idea there?

Answer 11

.8^9 *.2

Answer 12

?

Answer 13

almost. take what you have there and multiply by 10. There are 10 possibilities for the game that you are below 200. The 1st game, 2nd game, ... , 10th game.

Answer 14

.8^9 * 2?

Answer 15

10 * .8^9 * 2

Answer 16

Final answer: .8^10 + 10 * .8^9 * 2 = whatever that is

Answer 17

thank you!!!

Answer 18

means of course
\[10\times (.8)^9\times .2\]