A box of 11 transistors has 4 defective ones. A) If 2 transistors are drawn from the box together, what is the probability that both transistors are defective?
B) If 2 transistors are drawn from the box together, what is the probability that neither transistor is defective?
C) If 2 transistors are drawn from the box together, what is the probability that one transistor is defective?

(a) 0.1325

(b) 0.4045

(c) 0.463

Step-by-step explanation:

Let X denote the number of defective transistors.

The proportion of defective transistors is, p = 4/11 = 0.364.

All the transistors are independent of the others.

The random variable X follows a binomial distribution.

(a)

Compute the probability that both transistors are defective, if 2 transistors are drawn from the box together as follows:

(b)

Compute the probability that neither transistors are defective, if 2 transistors are drawn from the box together as follows:

(c)

Compute the probability that one transistors are defective, if 2 transistors are drawn from the box together as follows:

55% are girls and 45% are boys : ) hope this

step-by-step explanation:

substitute,

circle 1 = 5

circle 2 = 20

circle 3 = 44

in this case no matter how you add, you will always get a perfect square..

hence proved

happy to

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