Chapter 2 Section 2 Question 6
Question
Suppose a standard deck of 52 playing cards is thoroughly shuffled and a single card is drawn. Suppose an ace has value 1, a jack has value 11, a queen has value 12, and a king has value 13.
  1. Compute $P(X = x)$ for every real number x, when X is the value of the card drawn.
  2. Suppose that Y = 1, 2, 3, or 4 when a diamond, heart, club, or spade is drawn. Compute P(Y = y) for every real number y.
  3. 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, when X is the value of the card drawn.
$$P(X=1)=\frac{4}{52}\approx 0.077$$ $$P(X=11)=\frac{4}{52}\approx 0.077$$ $$P(X=12)=\frac{4}{52}\approx 0.077$$ $$P(X=13)=\frac{4}{52}\approx 0.077$$ $P(X=x)=0$ for all $x\notin\{1,11,12,13\}$
(b) Suppose that Y = 1, 2, 3, or 4 when a diamond, heart, club, or spade is drawn. Compute P(Y = y) for every real number y.
$$P(Y=1)=\frac{13}{52}=0.25$$ $$P(Y=2)=\frac{13}{52}=0.25$$ $$P(Y=3)=\frac{13}{52}=0.25$$ $$P(Y=4)=\frac{13}{52}=0.25$$ $P(Y=y)=0$ for all $y\notin\{1,2,3,4\}$
(c) Compute $P(W = w)$ for every real number w when $W = X + Y$.
Probability that X,Y add to 2: $$P(W=2)=P(X=1)*P(Y=1)$$ $$P(W=2)=0.077*0.25$$ $$P(W=2)\approx 0.019$$ Probability that X,Y add to 3: $$P(W=3)=P(X=1)*P(Y=2)$$ $$P(W=3)=0.077*0.25$$ $$P(W=3)\approx 0.019$$ Probability that X,Y add to 4: $$P(W=4)=P(X=1)*P(Y=3)$$ $$P(W=4)=0.077*0.25$$ $$P(W=4)\approx 0.019$$ Probability that X,Y add to 5: $$P(W=5)=P(X=1)*P(Y=1)$$ $$P(W=5)=0.077*0.25$$ $$P(W=5)\approx 0.019$$ Probability that X,Y add to 12: $$P(W=12)=P(X=11)*P(Y=1)$$ $$P(W=12)=0.077*0.25$$ $$P(W=12)\approx 0.019$$ Probability that X,Y add to 13: $$P(W=13)=P(X=11)*P(Y=2)+P(X=12)*P(Y=1)$$ $$P(W=13)=0.077*0.25+0.077*0.25$$ $$P(W=13)\approx 0.039$$ Probability that X,Y add to 14: $$P(W=14)=P(X=11)*P(Y=3)+P(X=12)*P(Y=2)+P(X=13)*P(Y=1)$$ $$P(W=14)=0.077*0.25+0.077*0.25+0.077*0.25$$ $$P(W=14)\approx 0.058$$ Probability that X,Y add to 15: $$P(W=15)=P(X=11)*P(Y=4)+P(X=12)*P(Y=3)+P(X=13)*P(Y=2)$$ $$P(W=15)=0.077*0.25+0.077*0.25+0.077*0.25$$ $$P(W=15)\approx 0.058$$ Probability that X,Y add to 16: $$P(W=16)=P(X=12)*P(Y=4)+P(X=13)*P(Y=3)$$ $$P(W=16)=0.077*0.25+0.077*0.25$$ $$P(W=16)\approx 0.039$$ Probability that X,Y add to 17: $$P(W=17)=P(X=13)*P(Y=4)$$ $$P(W=17)=0.077$$ $$P(W=17)\approx 0.019$$ $P(W=w)=0$ for all $w\notin\{2,3,4,5,12,13,14,15,16,17\}$