Question

A manufacturer of metal pistons finds that on the average, 12% of his pistons are rejected because they are either oversize or undersize. What is the probability that a batch of 10 pistons will contain at least 2 rejects?

Answer 1

In a binomial distribution

n = 10

p = `12/100 = 3/25`

q =  – p = `1 - 3/25` q = = `22/25`

PX = x) = nc_xp^xq^n-x 

P(at least 2 rejects) = P(X ≥ 2)

= `1 - "P"("X" < 2)`

= `1 - ["P"("X" = 0) + "P"("X" = 1)]`

= `1 - [(22)^10/(25)^10 + (30 xx (22)^9)/(25)^10]`

= `1 - {(22)^9/(25)^10 [22 + 30]}`

= `1 - {(52 xx (22)^9)/(25)^10}`

= `1 - {(52 xx 1.207 xx 10^12)/(9.528 xx 10^13)}`

= `1 - [(52 xx 1.207)/(9.528 xx 10)]`

= ` -[62.764/95.28]`

= `1 - 0.6587`

= 0.3413