A child has a box of crayons containing 16 crayons. 4 of these crayons are damag

Question

A child has a box of crayons containing 16 crayons. 4 of these crayons are damaged. Suppose the child randomly selects 2 of the 16 caryons to use.(without replacement). Let X equal the number of damaged crayons selected. Calculate the probability that x= 0. Calculate the probability that x= 1. Calculate the probability that x= 2. Calculate the expected value of X. Calculate the variance of X. Calculate the variance of 5X - 27. In how many ways can 2 crayons be selected from 16 crayons where order does not matter? We pick 2 crayons. In how many ways can 1 be damaged and 1 not damaged.

Explanation / Answer

Total number of ways of selecting 2 crayons out of 16 = 16C2 = 120

a. P(x=0) = number of ways of selecting 2 crayons out of 12 non-damaged/Total number of ways of selecting 2 crayons out of 16

= 12C2/16C2 = 66/120= 11/20 = 0.55

b. P(x=1) = number of ways of selecting 1 crayons out of 12 non-damaged and another out of 4 damaged/Total number of ways of selecting 2 crayons out of 16

= 12C1*4C1/16C2 = 12*4/120= 2/5 = 0.40

c. P(x=2) = number of ways of selecting 2 out of 4 damaged/Total number of ways of selecting 2 crayons out of 16

= 4C2/16C2 = 6/120= 1/20 = 0.05

d. E(X) = 0*0.55 + 1*0.40 + 2*0.05 = 0.5

e. Variance = (0-0.5)^2*0.55 + (1-0.5)^2*0.40 + (2-0.5)^2*0.05 = 0.35

f. Variance(5X-27) = 25*Variance(X) = 25*0.35 = 8.75

g. Total number of ways of selecting 2 crayons out of 16 = 16C2 = 120

h. 4C1*12C1 = 12*4 = 48

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