Probability & Statistics Ch.2 Section 2 Question 5

Solutions for “Probability and Statistics: The Science of Uncertainty” (Second Edition). These are solutions I have come up with; I offer no guarantee of accuracy.

Question

Suppose that a bowl contains 100 chips: 30 are labelled 1, 20 are labelled 2, and 50 are labelled 3. The chips are thoroughly mixed, a chip is drawn, and the number X on the chip is noted.

a) Compute $P(X = x)$ for every real number x. b) Suppose the first chip is replaced, a second chip is drawn, and the number Y on the chip noted. Compute $P(Y = y)$ for every real number y. c) Compute $P(W = w)$ for every real number w when $W = X + Y$ .

Solution

(a) Compute $P(X = x)$ for every real number x

P(X=1)=\frac{30}{100}=0.3

P(X=2)=\frac{20}{100}=0.2

P(X=3)=\frac{50}{100}=0.5

$P(X=x)=0$ for all $x\notin\{1,2,3\}$

(b) Compute $P(Y = y)$ for every real number y

Since the first chip is replaced before drawing the second:

P(Y=1)=\frac{30}{100}=0.3

P(Y=2)=\frac{20}{100}=0.2

P(Y=3)=\frac{50}{100}=0.5

$P(Y=y)=0$ for all $y\notin\{1,2,3\}$

Note: This answer is exactly the same as part (a) because of replacement.

(c) Compute $P(W = w)$ for every real number w when $W = X + Y$

Probability that X + Y = 2:

P(W=2)=P(X=1) \times P(Y=1) = 0.3 \times 0.3 = 0.09

Probability that X + Y = 3:

P(W=3)=P(X=1) \times P(Y=2)+P(X=2) \times P(Y=1) = 0.3 \times 0.2 + 0.2 \times 0.3 = 0.12

Probability that X + Y = 4:

P(W=4)=P(X=1) \times P(Y=3)+P(X=2) \times P(Y=2)+P(X=3) \times P(Y=1)

= 0.3 \times 0.5 + 0.2 \times 0.2 + 0.5 \times 0.3 = 0.34

Probability that X + Y = 5:

P(W=5)=P(X=2) \times P(Y=3)+P(X=3) \times P(Y=2) = 0.2 \times 0.5 + 0.5 \times 0.2 = 0.2

Probability that X + Y = 6:

P(W=6)=P(X=3) \times P(Y=3) = 0.5 \times 0.5 = 0.25

$P(W=w)=0$ for all $w\notin\{2,3,4,5,6\}$