Question

a) The probability of a patient recovering from a rare blood disease is 0.4. If it is known that 100 people have contracted this disease, what is the probability that more than 46 will survive?

b) Suppose that the probability of a 25-year-old man wanting to celebrate his fiftieth birthday is 74.2% and the probability of a 22-year-old woman living to her forty-seventh birthday is 0.902. If both people get married this year, what are the chances of that couple living to celebrate their silver wedding?

Answer 1

a) P( Recovring from disease ) =0.4 = p

100 people have contracted this disease, n= 100

Let X be the number of people recovering from this disease

X~ Binomial ( 100, 0.4)

Since , sample size is large, so we will use normal approximation

X~ Normal ( np, np(1-p))

X~ Normal ( 40, 24)

P( more than 46 will survive) = P ( X > 46)

= P( > )

= P( z > 1.22)

= 1- P(z < 1.22)

= 1- 0.88877

= 0.11123

2) P( 25-year-old man wanting to celebrate his fiftieth birthday ) = 0.742 =P( 25 year old man living for more 25 yrs)

P (22-year-old woman living to her forty-seventh birthday ) = 0.902 = P( 22 yr old woman living for more 25 yrs)

P( couple living to celebrate their silver wedding)

= P( 25 year old man living for more 25 yrs) *P( 22 yr old woman living for more 25 yrs)

= 0.742 * 0.902

= 0.669284