Question

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1) A bowl contains eleven bananas, five of which are ripe. If three bananas are selected at random without replacement, find the following probabilities:

Exactly two are ripe

At most one is ripe

2) Experiment: (Binomial Distribution) In a survey, 83% of children recognized the character SpongeBob Square Pants. If eight children are selected at random, find the probability:

Exactly three will recognize SpongeBob.

Find the mean for this distribution

Find the variance for this distribution.

Answer 1

1) A bowl contains eleven bananas, five of which are ripe. If three bananas are selected at random without replacement, find the following probabilities:

P[ Exactly two are ripe ] = P[ Ripe banana chosen ]*P[ Ripe banana chosen | first banana was ripe ]*P[ good banana chosen | after two ripe bananas ]*Number of combinations

P[ Ripe banana chosen ] = 5/11

P[ Ripe banana chosen | first banana was ripe ] = 4/10

P[ good banana chosen ] = ( 11 - 5 )/9 = 6/9

Number of combinations = 3

P[ Exactly two are ripe ] = 3*(5/11)*(4/10)*(6/9)

P[ Exactly two are ripe ] = 360/ 990

P[ Exactly two are ripe ] = 4/11

P[ Exactly two are ripe ] = 0.3636

P[ At most one is ripe ] = 1 - P[ Exactly two are ripe ] - P[ all three are ripe ]

P[ all three are ripe ] = P[ Ripe banana chosen ]*P[ Ripe banana chosen | first banana was ripe ]*P[ ripe banana chosen | after two ripe bananas ]

P[ all three are ripe ] = (5/11)*(4/10)*(3*9)

P[ all three are ripe ] = 60/ 990

P[ all three are ripe ] =2/33

P[ all three are ripe ] = 0.0606

P[ At most one is ripe ] = 1 - 0.3636 - 0.0606

P[ At most one is ripe ] = 0.5758

In a survey, 83% of children recognized the character Sponge Bob Square Pants. If eight children are selected at random

The population proportion of success is p = 0.83 , also, 1 - p = 1 - 0.83 = 0.17 , and the sample size is n= 8 . We need to compute Pr(X=3)

This implies that

Mean = n*p

Mean = 8*0.83

Mean = 6.64

Variance = n*p*(1-p)

Variance = 8*0.83*0.17

Variance = 1.1288