Question 316060: a basketball player gets a basket on the average of once every 10 times he shoots. if this player shoots 6 times, what is the probability that
(round answers to the 3 decimal):
a) exactly 3 shots go in the basket?
P of success= .10 P of not success= .90
k= 3 shots; n= 6
so:
b(6,3; .10) = (6,3)(.10)^3(.90)^3 = 6!/3!3!= 20(.001)(.729)= 0.013
P(exactly 3 shots)=0.013
b) none of the shots go in?
b(6,0; .10) = (6,0)(.10)^0(.90)^6 = 6!/6!0!= 1(0)(.5314)=0.531
P(exactly 0 shots)=0.531
c) less than four shots go in the basket?
b(6,3;.10) + b(6,2;.10) + b(6,1;.10) + b(6,0;.10)=
0.013 + 6!/4!2!=15(.01)(.6561)=0.98415 + 6!/5!1!=6(.10)(.59049)=.354294 +.531=
0.013+0.98415+.59049+.531= .996709
P(less than 4 shots)=.997
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A basketball player gets a basket on the average of once every 10 times he shoots. if this player shoots 6 times, what is the probability that
-----------------------------------------
a) exactly 3 shots go in the basket?
P of success= .10 P of not success= .90
k= 3 shots; n= 6
so:
b(6,3; .10) = (6,3)(.10)^3(.90)^3 = 6!/3!3!= 20(.001)(.729)= 0.0146
P(exactly 3 shots)=0.0146
-----------------------------------
b) none of the shots go in?
b(6,0; .10) = (6,0)(.10)^0(.90)^6 = 6!/6!0!= 1(0)(.5314)=0.531
P(exactly 0 shots)=0.531
-----------------------------------
c) less than four shots go in the basket?
b(6,3;.10) + b(6,2;.10) + b(6,1;.10) + b(6,0;.10)=
0.013 + 6!/4!2!=15(.01)(.6561)=0.98415 + 6!/5!1!=6(.10)(.59049)=.354294 +.531=
0.013+0.98415+.59049+.531= .9987
P(less than 4 shots)=.9987
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Cheers,
Stan H.
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