Question

64. A batch of 50 different automatic typewriters contains exactly 10 defective machines. What is the probability of finding (a) At least one defective machine in a random group of five machines? (b) At least two defective machines in a random group of 10 machines? 4 Chapter 5 General Counting Methods for Arrangements and Selections (c) The first defective machine to be the kth machine taken apart for inspection in a random sequence of machines? (d) The last defective machine to be the kth machine taken apart?

Answer 1

10:20 Answer: P= 10 = 10:20] (a) Given n=5 let Hie random variable x denotes member of defective machines in a group of 5 machines. let the random variable x denotes number of defective machines a group of 5 machines. Clearly x follows a Binomial distribution With parameters (n=5, P-0.20) The probability mass function offis defined as belo hi P(x=)= (1.) pour compae ; «=0,1,2---1 - () X0-20*80 86*** ;x=0,1,2,3,4,5 The required probability io givenas PCX2101- Pxzo) PCX31)=1-13]x0.20*x0.805 -1-0.3277 = 10-6783]

(b) D=10 Clearly, x follows Binomial distribution bith parameters (n=10,10 =0-20) PCX = ) = (12] pe C1-P) *)=2; x =0,1,2--1 PCx=~) - (10 7x0.20* x 0.8010-8; 2=0,1,2 --10 The required probability is givenay P(X22)-1-PCX51) - 1- [PCx=0)+ PCx = ] -1-(1) X0-2080.800+('978020 x0.98] 31-10.1074 +0.2684] 21-0.3758 = 0.6242 ce) The probability mass function of x is defined as below: p(x=2) = (1-P) &-p; a=1,2... = 0-804m!X0-20; 2=1,2

T40 P Therefore, the probability of finding the k machine to the first defective in a random sequence will be be PCX = k) = 0.8k-1X0.20) d) clearly x follows a negative binomial distribution With parameters (r=10, P=0.20) The probability mass function of x isdefined 08' below. Plx=4) (-Co-pje opt just, +1,442-- - [:-) +o.gle x o °, 140, 1, 19 -- Therefore, the probability of finding the kth machine to be post defectue in a random Sequence Wi Ato be P(X=K) = (x + y)20.921040.210 1 -10