Question

1. cece and owen are visiting the library with their mother. based on their previous visits, the distributions of the number of books they will want to check out are given in the following tables.

cece

number of books	0	1	2	3
probability	0.05	0.20	0.45	0.30

owen

number of books	0	1	2
probability	0.10	0.35	0.55

a. assuming they make their decisions independently, what is the probability that cece wants to check out 1 book and owen wants to check out 1 book? what other ways would lead them to want to check out 2 books total?

b. what is the probability that the total number of books cece and owen want to check out is 2? show your work.

c. let t = the total number of books that cece and owen want to check out. complete the probability distribution table for t.

t	0	1	2	3	4	5
probability				0.2975	0.3525	0.165

d. find the mean of the distribution t.

Explanation:

Step1: Calculate probability for part a

Since they make independent decisions, use the formula \(P(A\cap B)=P(A)\times P(B)\). For Cece checking out 1 book (\(P(C = 1)=0.20\)) and Owen checking out 1 book (\(P(O = 1)=0.35\)), the probability is \(0.20\times0.35 = 0.07\).
Other ways to get a total of 2 books: Cece 0 and Owen 2 (\(P(C = 0)\times P(O = 2)=0.05\times0.55 = 0.0275\)) and Cece 2 and Owen 0 (\(P(C = 2)\times P(O = 0)=0.45\times0.10 = 0.045\))

Step2: Calculate probability for part b

The probability that the total number of books is 2 is the sum of the probabilities of the three cases (Cece 1 - Owen 1, Cece 0 - Owen 2, Cece 2 - Owen 0). So \(P(T = 2)=0.07+0.0275 + 0.045=0.1425\)

Step3: Calculate probability for part c (for \(T = 0\))

\(P(T=0)=P(C = 0)\times P(O = 0)=0.05\times0.10 = 0.005\)
For \(T = 1\): \(P(C = 0)\times P(O = 1)+P(C = 1)\times P(O = 0)=0.05\times0.35+0.20\times0.10=0.0175 + 0.02=0.0375\)

Step4: Calculate the mean for part d

The formula for the mean \(\mu=\sum_{i}t_{i}P(t_{i})\)
\(\mu=0\times0.005+1\times0.0375 + 2\times0.1425+3\times0.2975+4\times0.3525+5\times0.165\)
\(=0 + 0.0375+0.285+0.8925+1.41+0.825\)
\(=3.45\)

Answer:

a. The probability that Cece wants to check out 1 book and Owen wants to check out 1 book is \(0.07\). Other ways: Cece 0 - Owen 2 (\(P = 0.0275\)) and Cece 2 - Owen 0 (\(P = 0.045\))
b. \(P(T = 2)=0.1425\)
c. For \(T = 0\), probability \(=0.005\); for \(T = 1\), probability \(=0.0375\)
d. The mean of the distribution \(T\) is \(3.45\)