It is determined while using the binomial distribution that there is still a 1.145=114.5% chance that she produces no more than 11 of them.
<h2>Calculating the probability</h2>There are just two possible results for each throw. Either she succeeds or she fails. The binomial probability distribution is employed to answer this issue since the probability of completing a shot is regardless of all other throws.
Binomial probability distribution-
where,
the no. of success= x
the no. of trials = n
the probability of a success on one trial = p
The probability of throwing not more than 11 will be:
P(X<11) = P(X=0) + P(X=1) +P(X=2) + P(X=3) +P(X=4) +P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)+P(X=11)
Where,
≈0
≈0
≈0
≈0
≈0
= 0.0003
=0.0016
=0.0062
=0.0188
=0.0463
=0.9232
=0.1488
So,
P(X<11) = P(X=0) + P(X=1) +P(X=2) + P(X=3) +P(X=4) +P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)+P(X=11)
=0+0+0+0+0+0.0003+0.0016+0.0062+0.0188+0.0463+0.9232+0.1488 =1.145
Therefore, she makes 1.145=114.5% probability, no more than 11 of them.
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