If there are known to be 4 broken transistors in a box of 12, and 3 transistors are drawn at random, what is the probability that none of the 3 is broken?

Options

A. 0.25
B. 0.255
C. 0.375
D. 0.556

Answer

B

Explanation

Since there are 4 broken transistors, there must be 8 good ones. P(first pick is good)=`8/(12)`. Of the remaining 11 transistors, 7 are good, and so P(second pick is good)`=7/(11)`. Finally, P(third pick is good )=`6/(10)`. Therefore, P(all three are good)=`8/(12)*7/(11)*6/(10)*(14)/(55)~~0.255`. [22]