Question

A box of 12 transistors has 3 defective ones. If 2 transistors are drawn from the box together, what is the probability that

(I)both transistors are defective?

(II) neither transistor is defective?

(III)one transistor is defective?

Answer 1

Solution:

Here binomial(n,k) is binomial coefficient (ⁿ_k)=n!/k!/(n-k)!

(i) P(both transistors are defective)=m/n

m=binomial(3,2)=3!/2!/1!=3,

n=binomial(12,2)=12!/2!/10!=66.

P(both transistors are defective)=3/66=1/22,

(ii)P(neither transistor is defective)=m/n

m=binomial(9,2)=9!/2!/7!=36,

n=binomial(12,2)=12!/2!/10!=66.

P(neither transistor is defective)=36/66=6/11,

(iii) P(one transistor is defective)=

=1-P(both transistors are defective)-P(neither transistor is defective)=

=1-1/22-6/11=9/22.

Answer:

P(both transistors are defective)=1/22,

P(neither transistor is defective)=6/11,

P(one transistor is defective)=9/22.