Question

21.5% of flowers of a certain species bloom "early" (before May 1st). You work for an arboretum and have a display of these flowers

a) In a row of 35 flowers, what is the probability that 8 will bloom early?

b) In a row of 35 flowers, what is the probability that fewer than 9 will bloom early?

c) As you walk down a row of these flowers, how many flowers do you expect to have to observe (on average) in order to see the first one that blooms early? (Keep your answer as a decimal.)

d) In a row of 50 flowers, what is the probability that more than 8 will bloom early?

e) In a row of 50 flowers, what is the probability that between 8 and 14 (inclusive) will bloom early?

f) In a row of flowers, what is the probability that you will have to observe 6 flowers in order to see the first one that blooms early?

g) In a row of flowers, what is the probability that you will observe more than 7 flowers to see the first one that blooms early?

Answer 1

a) Let X be the number of flowers that bloom early

X follow Binomial with n= 35 , p=0.215

Probability mass function of a Binomial distribution  is

, x=0,1,...n

In a row of 35 flowers , probability that 8 will bloom early =0.1559

b) To find P( X < 9)

= 0.6677

In a row of 35 flowers , probability that less than 9  will bloom early =0.6677

Note : Used excel formula " =BINOM.DIST(8,35,0.215,cumulative)

c) Let N be the number of flowers observed to see the first one that blooms early

N follow Geometric distribution with p =0.215

P( N=n) = ( 1- p)^n-1 p

expectation of N = 1/p =1/0.215 = 4.7

Expected number of flowers to observe = 4.7

d) Let X be the number of flowers that bloom early

X follow Binomial with n= 50 , p=0.215

To find P( X > 8 )

= 1-0.2234

= 0.7766

In a row of 50 flowers , probability that more than 8 will bloom early =0.7766

Note : Used excel formula " =BINOM.DIST(8,50,0.215,cumulative)

e) To find

= 0.7695

In a row of 50 flowers , probability that between 8 and 14 will bloom early =0.7695

Note : Used excel formula " =BINOM.DIST(14,50,0.215,cumulative)-=BINOM.DIST(7,50,0.215,cumulative)"

We can calculate manually too.

f) Let N be the number of flowers observed to see the first one that blooms early

N follow Geometric distribution with p =0.215

P( N=n) = ( 1- p)^n-1 p

P( N=6) = (1-0.215) ⁵*0.215 = 0.0641

Probability that 6 flowers are to observed = 0.0641

g) To find P( N> 7)

P( N> 7) = 1-P( N 7 )

= 1- (P(N=1) +....P( N=7) )

= 1-0.8163

= 0.1837

Probability that more than 7  flowers are to observed = 0.1837