Question

In: Statistics and Probability

A earton of 12 transistor batteries contains 4 that are defective. In how many different ways...

A earton of 12 transistor batteries contains 4 that are defective. In how many different ways can one choose 3 of these batteries so that
a) none of the defective batteries is included
b) exactly 1 of the defective batteries is included
c) exactly 2 of the defective batteries are included
d) exactly 3 of the defective batteries are included

Expert Solution

total number of transistor batteries = 12

number of defective batteries = 4

number of good batteries = (12-4) = 8

a).number of ways none of the defective batteries is included is:-

[ we will select all the 3 batteries from 8 good batteries ]

b).number of ways exactly 1 of the defective batteries is included is:-

[ select 1 defective batteries from 4 defective batteries and rest (3-1) = 2 batteries from 8 good batteries ]

c).number of ways exactly 2 of the defective batteries is included is:-

[ select 2 defective batteries from 4 defective batteries and rest (3-2) = 1 batteries from 8 good batteries ]

d).number of ways exactly 3 of the defective batteries is included is:-

[ select all the 3 defective batteries from 4 defective batteries ]

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orchestra answered 1 year ago

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