Question

the probaility of a pitcher pitching in the strike zone is 95%. what is the probability that the pitcher will get at least 2 balls in the strike zone in the next 3 pitches?

Answer 1

1-(0.95x0.05x0.05 + 0.05^3) ?

Answer 2

0.95^2x0.05 + 0.95^3?

Answer 3

\[P(2inStrikeZone)=\left(\begin{matrix}3 \\ 2\end{matrix}\right)\times (0.95)^{2}\times (1-0.95)=3\times (0.95)^{2}\times 0.05\]

Answer 4

thats 0.135 how can that be right..

Answer 5

bhi

Answer 6

\[P(1inStrikeZone)=\left(\begin{matrix}3 \\ 1\end{matrix}\right)\times 0.95\times (0.05)^{2}=0.007125\]
\[P(0inStrikeZone)=\left(\begin{matrix}3 \\ 0\end{matrix}\right)\times 0.95^{0}\times (0.05)^{3}=0.000125\]
P(2) + P(1) + P(0) = P(up to 2 in strike zone) = 0.135375 + 0.007125 + 0.000125
= 0.142625
\[P(3inStrikeZone)=\left(\begin{matrix}3 \\ 3\end{matrix}\right)\times (0.95)^{3}=0.857375\]

Answer 7

Note that the highest probability is of getting 3 in the strike zone (0.857375). If the probability of getting up to 2 in the strike zone (0.142625 is added to the probability of getting 3 in the strike zone (0.857375) the result is exactly 1, confirming the validity of the calculations.

Answer 8

@powerangers69 Do you understand the explanation?

Answer 9

is 3 1 3c1?

Answer 10

i no understand

Answer 11

I used the binomial distribution to solve this problem. Do you know how to use the binomial distribution?

Answer 12

nope sorry

Answer 13

http://en.wikipedia.org/wiki/Binomial_distribution

Answer 14

can do without binomail distr?

Answer 15

Yes, this problem can be solved very easily.
The probability of getting all 3 of the next three balls in the strike zone is:
\[0.95\times 0.95\times 0.95=0.857375\]
All the possible outcomes of the next 3 balls getting in the strike zone are:
0 balls, 1 ball, 2 balls and 3 balls.
P(0) + P(1) + P(2) + P(3) = 1
We already found that P(3) = 0.857375,
therefore P(0) + P(1) + P(2) = 1 - P(3) = 1 - 0.857375 = 0.142625