Expected value of a discrete random variable Problem set B Worked solutions

The table shows the probability distribution of X, the number of pets in a household.

1. (a) Find E(X).

   • \(E(X)=0\times0.22 + 1\times0.31 + 2\times0.27 + 3\times0.12 + 4\times0.08=1.53\)

2. (b) Can X take the value E(X)? Explain.

   • X counts pets, so it is a whole number. E(X) is the mean over many households, not one household.

A quiz round awards Z points. The table shows the probability distribution of Z. Find E(Z).

• \(E(Z)=5\times0.45 + 10\times0.30 + 25\times0.20 + 50\times0.05=12.75\)
• E(Z) is the mean of Z over many trials.

The random variable Y has distribution \(P(Y=y)=0.02(y^2-10y+24)\) for y = 0, 1, 2, 3.

1. (a) Write the probability distribution of Y as a table.

   • \(P(Y=0)=0.02(0+0+24)=0.48\)
   • \(P(Y=1)=0.02(1-10+24)=0.3\)
   • \(P(Y=2)=0.02(4-20+24)=0.16\)
   • \(P(Y=3)=0.02(9-30+24)=0.06\)
   • The probabilities add to 1.

2. (b) Find E(Y).

   • \(E(Y)=0\times0.48 + 1\times0.3 + 2\times0.16 + 3\times0.06=0.8\)

The table shows the probability distribution of Y.

1. (a) Find the value of k.

   • \(0.06 + 0.04+k+2k=1\)
   • \(3k=0.9\)
   • \(k=0.3\)

2. (b) Find P(Y ≤ 7).

   • \(P(Y\le7)=0.06+0.04+0.3=0.4\)

3. (c) Find E(Y).

   • \(E(Y)=2\times0.06 + 4\times0.04 + 7\times0.3 + 10\times0.6=8.38\)

The table shows the probability distribution of Z, where k is a constant.

1. (a) Find the value of k.

   • \(0.08+k+0.22+3k+0.3=1\)
   • \(0.6+4k=1\)
   • \(k=0.1\)

2. (b) Find P(Z ≤ 9).

   • \(P(Z\le9)=0.08+0.1+0.22=0.4\)

3. (c) Find P(Z > 6 | Z ≤ 12).

   • \(P(Z>6\mid Z\le12)=\frac{P(6<Z\le12)}{P(Z\le12)}=\frac{0.52}{0.7}=0.743\)

A shelf holds 6 hardback books and 3 paperback books. Two books are taken at random without replacement. X is the number of paperbacks taken.

1. (a) Find the probability distribution of X.

   • \(P(X=0)=\frac{6}{9}\times\frac{5}{8}=\frac{5}{12}\)
   • \(P(X=1)=2\times\frac{3}{9}\times\frac{6}{8}=\frac{1}{2}\)
   • \(P(X=2)=\frac{3}{9}\times\frac{2}{8}=\frac{1}{12}\)

2. (b) Find E(X).

   • \(E(X)=0\times\frac{5}{12}+1\times\frac{1}{2}+2\times\frac{1}{12}=\frac{2}{3}\)
   • Check: 2 × 3/9 = 2/3.

A kiosk records X, the number of drinks a customer buys. The table shows its probability distribution.

1. (a) Find E(X).

   • \(E(X)=0\times0.20 + 1\times0.40 + 2\times0.30 + 3\times0.10=1.3\)

2. (b) 150 customers are expected in a day. How many drinks should the shop expect to sell?

   • \(150\times1.3=195\)
   • Multiply the mean for one customer by the number of customers.

The number of books, Y, borrowed by a student in a visit has distribution \(P(Y=y)=0.01(y^2-y+15)\) for y = 2, 3, 4, 5.

1. (a) Write the probability distribution of Y as a table.

   • \(P(Y=2)=0.01(4-2+15)=0.17\)
   • \(P(Y=3)=0.01(9-3+15)=0.21\)
   • \(P(Y=4)=0.01(16-4+15)=0.27\)
   • \(P(Y=5)=0.01(25-5+15)=0.35\)
   • The probabilities add to 1.

2. (b) Find E(Y).

   • \(E(Y)=2\times0.17 + 3\times0.21 + 4\times0.27 + 5\times0.35=3.8\)

A box holds 9 fuses, 3 of them faulty. Three fuses are tested at random without replacement. X is the number of faulty fuses tested.

1. (a) Find the probability distribution of X.

   • \(P(X=0)=\frac{6}{9}\times\frac{5}{8}\times\frac{4}{7}=\frac{5}{21}\)
   • \(P(X=1)=3 \times \frac{3}{9}\times\frac{6}{8}\times\frac{5}{7}=\frac{15}{28}\)
   • \(P(X=2)=3 \times \frac{3}{9}\times\frac{2}{8}\times\frac{6}{7}=\frac{3}{14}\)
   • \(P(X=3)=\frac{3}{9}\times\frac{2}{8}\times\frac{1}{7}=\frac{1}{84}\)

2. (b) Find E(X).

   • \(E(X)=0\times\frac{5}{21}+1\times\frac{15}{28}+2\times\frac{3}{14}+3\times\frac{1}{84}=1\)
   • Check: 3 × 3/9 = 1.

The table shows the probability distribution of Y, the number of points a player scores in a round.

1. (a) Find the value of k.

   • \(k+0.1+2k+0.2+0.1=1\)
   • \(3k=0.6\)
   • \(k=0.2\)

2. (b) Find E(Y).

   • \(E(Y)=1\times0.2 + 2\times0.1 + 3\times0.4 + 4\times0.2 + 5\times0.1=2.9\)

3. (c) Find P(Y > E(Y)).

   • E(Y) = 2.9, so Y > 2.9 means y = 3, 4, 5.
   • \(P(Y>E(Y))=0.4+0.2+0.1=0.7\)