In a shipment of 15 sets of golf clubs, 3 are lefthanded. If 4 sets of golf club

Question

In a shipment of 15 sets of golf clubs, 3 are lefthanded. If 4 sets of golf clubs are selected, what is the probability that

a. exactly 1 is left-handed?

b. at least 1 is left-handed?

c. no more than 2 are left-handed?

d. What is the mean number of left-handed sets of golf clubs that you would expect to find in the sample of 4 sets of golf clubs.

The professor made some change of this problem so PLEASE NOTICE:

change the mean clubs to 6.0. The constant is still e=2.7183. Complete a, b, c, and a new "e". "e" between 2 and 5 inclusive. Use the Basic method, the Lambda Chart. method, and the Table.

Explanation / Answer

we solve this problem by using binomial distribution

n=4 ; p = 3/15 = 1/5 =0.2 ; q= 1-p = 4/5 =0.8 {Given Parameter

a)exactly 1 is left-handed

here we haveto calculate prob of exactly one lefthanded and other 3 are Right handed person.

pr(X=1)= 4c1*0.2*0.8^3

Pr(X=1)=0.4096

b) At least 1 is left handed

Pr(X=1)+Pr(X=2)+Pr(X=3)

=1-Pr(X=0)

=nCr*(p^r)*(q^n-r)

=1 - 4C0*(0.2^0)*(0.8^4)

=0.5904

c)no more than 2 are left-handed

=Pr(X=0)+Pr(X=1)+Pr(X=2)

=4C0*(0.2^0)*(0.8^4) +(4C1*(0.2^1)*(0.8^3)) +(4C2*(0.2^2)*(0.8^2))

=0.4096 + 0.4096 + 0.1536

=0.9728

d)for binomial distribution mean =np

n=4 ; p= 0.2

=4*0.2

=0.8

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