Question

1) If the boys and girls have the same probabilities of being born. What is the probability that a family of 5 children selected randomly has atleast one boy?

2) Of the pieces produced by a machine in particular, 0.8% are defective. If an aleatory sample of 8 pieces produced by this machine contains two or more defective pieces, the machine will be turned off to make reparations. Find the probability that the machine would turn off for reparations based on the sampling plan.

Answer 1

1)

The boys and girls have the same probabilities of being born.

So, P( boys ) = P( Girls ) = p = 0.5

Let , X be the Number of boys in sample of 5 children.

X follows Binomial distribution with n = 5 and p = 0.5

We have to find P( X >= 1 )

P( X >= 1 ) = 1 - P( X < 1 ) = 1 - P( X <= 0 )

Using Excel function ,   =BINOMDIST( x , n , p, 1 )

P( X <= 0 ) = BINOMDIST( 0, 5, 0.5, 1 ) = 0.03125

So , P( X >= 1 ) = 1 - 0.03125 = 0.96875

The probability that a family of 5 children selected randomly has atleast one boy is 0.96875

2)

Let, X be the number of defective pieces in a sample.

P( defective piece ) = 0.8% = 0.008

sample size n = 8

X follows Binomial distribution with n =8 and p = 0.008.

If an aleatory sample of 8 pieces produced by this machine contains two or more defective pieces, the machine will be turned off to make reparations

So we have to find P( X >= 2 )

P( X >= 2 ) = 1 - P( X < 2 ) = 1 - P( X <= 1 )

Using Excel function ,   =BINOMDIST( x , n , p, 1 )

P( X <= 1) = BINOMDIST( 1, 8, 0.008, 1 ) = 0.998264

So, P( X >= 2 ) = 1 - 0.998264 = 0.001736

The probability that the machine would turn off for reparations based on the sampling plan is 0.001736 .