Chapter 2 Section 2 Question 5

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.

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

Answer

Author: Mohammad-Ali Bandzar| Date:Oct 14 2020

- (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) 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.
- $$P(Y=1)=\frac{30}{100}=0.3$$ $$P(Y=2)=\frac{20}{100}=0.2$$ $$P(Y=3)=\frac{50}{100}=0.5$$ $P(Y=y)=0$ for all$y\notin\{1,2,3\}$

Note: this answer is exactly the same as question a - (c) Compute $P(W = w$) for every real number w when $W = X + Y$.
- Probability that X,Y chips add to two: $$P(W=2)=P(X=1)*P(Y=1)$$ $$P(W=2)=(0.3)^2$$ $$P(W=2)=0.09$$ Probability that X,Y chips add to three: $$P(W=3)=P(X=1)*P(Y=2)+P(X=2)*P(Y=1)$$ $$P(W=3)=0.3*0.2+0.3*0.2$$ $$P(W=3)=0.12$$ Probability that X,Y chips add to four: $$P(W=4)=P(X=1)*P(Y=3)+P(X=2)*P(Y=2)+P(X=3)*P(Y=1)$$ $$P(W=4)=0.3*0.5+0.2*0.2+0.5*0.3$$ $$P(W=4)=0.34$$ Probability that X,Y chips add to five: $$P(W=5)=P(X=2)*P(Y=3)+P(X=3)*P(Y=2)$$ $$P(W=5)=0.2*0.5+0.5*0.2$$ $$P(W=5)=0.2$$ Probability that X,Y chips add to six: $$P(W=6)=P(X=3)*P(Y=3)$$ $$P(W=5)=0.5*0.5$$ $$P(W=5)=0.25$$ $P(W=w)=0$ for all $w\notin\{2,3,4,5,6\}$