Question

A basketball player has a 0.689 probability of making a free throw. If the player shoots 18 free throws, what is the probability that she makes no more than 11 of them

Answer 1

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.

Calculating the probability

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-

$P(X=x) = C_{n,x} .p^{x}(1-p)^{n-x}$

$C_{n,x} = n!/x! (n-x)!$

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,

$P(X=x) = C_{n,x} .p^{x}(1-p)^{n-x}$

$P(X=0) = C_{18,0} .(0.689)^{0}(0.311)^{18}$≈0

$P(X=1) = C_{18,1} .(0.689)^{1}(0.311)^{17}$≈0

$P(X=2) = C_{18,2} .(0.689)^{2}(0.311)^{16}$≈0

$P(X=3) = C_{18,3} .(0.689)^{3}(0.311)^{15}$≈0

$P(X=4) = C_{18,4} .(0.689)^{4}(0.311)^{14}$≈0

$P(X=5) = C_{18,5} .(0.689)^{5}(0.311)^{13}$= 0.0003

$P(X=6) = C_{18,4} .(0.689)^{6}(0.311)^{12}$=0.0016

$P(X=7) = C_{18,7} .(0.689)^{7}(0.311)^{11}$=0.0062

$P(X=8) = C_{18,8} .(0.689)^{8}(0.311)^{10}$=0.0188

$P(X=9) = C_{18,9} .(0.689)^{9}(0.311)^{9}$=0.0463

$P(X=10) = C_{18,10} .(0.689)^{10}(0.311)^{8}$=0.9232

$P(X=11) = C_{18,11} .(0.689)^{11}(0.311)^{7}$=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.

Learn more about probability here:

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