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

The table shows the probability distribution of X, the number of passengers in a taxi.

1. (a) Find E(X).

   • \(E(X)=1\times0.32 + 2\times0.28 + 3\times0.20 + 4\times0.12 + 5\times0.08=2.36\)

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

   • X counts passengers, so it is a whole number. E(X) is the mean over many taxis, not one taxi.

A spinner pays a prize, Z pounds. The table shows the probability distribution of Z. Find E(Z).

• \(E(Z)=10\times0.50 + 20\times0.30 + 50\times0.15 + 100\times0.05=23.5\)
• E(Z) is the mean of Z over many trials.

The random variable Y has distribution \(P(Y=y)=0.02(y^2-6y+18)\) for y = 0, 1, 2, 3.

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

   • \(P(Y=0)=0.02(0+0+18)=0.36\)
   • \(P(Y=1)=0.02(1-6+18)=0.26\)
   • \(P(Y=2)=0.02(4-12+18)=0.2\)
   • \(P(Y=3)=0.02(9-18+18)=0.18\)
   • The probabilities add to 1.

2. (b) Find E(Y).

   • \(E(Y)=0\times0.36 + 1\times0.26 + 2\times0.2 + 3\times0.18=1.2\)

The table shows the probability distribution of Y.

1. (a) Find the value of k.

   • \(0.45 + 0.25+k+2k=1\)
   • \(3k=0.3\)
   • \(k=0.1\)

2. (b) Find P(Y ≤ 5).

   • \(P(Y\le5)=0.45+0.25+0.1=0.8\)

3. (c) Find E(Y).

   • \(E(Y)=1\times0.45 + 3\times0.25 + 5\times0.1 + 8\times0.2=3.3\)

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

1. (a) Find the value of k.

   • \(0.12+k+0.28+3k+0.2=1\)
   • \(0.6+4k=1\)
   • \(k=0.1\)

2. (b) Find P(Z ≤ 6).

   • \(P(Z\le6)=0.12+0.1+0.28=0.5\)

3. (c) Find P(Z > 4 | Z ≤ 8).

   • \(P(Z>4\mid Z\le8)=\frac{P(4<Z\le8)}{P(Z\le8)}=\frac{0.58}{0.8}=0.725\)

A bag holds 5 cherry sweets and 7 lime sweets. Two sweets are taken at random without replacement. X is the number of cherry sweets taken.

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

   • \(P(X=0)=\frac{7}{12}\times\frac{6}{11}=\frac{7}{22}\)
   • \(P(X=1)=2\times\frac{5}{12}\times\frac{7}{11}=\frac{35}{66}\)
   • \(P(X=2)=\frac{5}{12}\times\frac{4}{11}=\frac{5}{33}\)

2. (b) Find E(X).

   • \(E(X)=0\times\frac{7}{22}+1\times\frac{35}{66}+2\times\frac{5}{33}=\frac{5}{6}\)
   • Check: 2 × 5/12 = 5/6.

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

1. (a) Find E(X).

   • \(E(X)=0\times0.15 + 1\times0.45 + 2\times0.30 + 3\times0.10=1.35\)

2. (b) 200 customers are expected in a day. How many loaves should the shop expect to sell?

   • \(200\times1.35=270\)
   • Multiply the mean for one customer by the number of customers.

The number of coffees, Y, bought by a customer has distribution \(P(Y=y)=0.01(y^2+3y+10)\) for y = 1, 2, 3, 4.

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

   • \(P(Y=1)=0.01(1+3+10)=0.14\)
   • \(P(Y=2)=0.01(4+6+10)=0.2\)
   • \(P(Y=3)=0.01(9+9+10)=0.28\)
   • \(P(Y=4)=0.01(16+12+10)=0.38\)
   • The probabilities add to 1.

2. (b) Find E(Y).

   • \(E(Y)=1\times0.14 + 2\times0.2 + 3\times0.28 + 4\times0.38=2.9\)

A drawer holds 6 black socks and 4 white socks. Three socks are taken at random without replacement. X is the number of white socks taken.

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

   • \(P(X=0)=\frac{6}{10}\times\frac{5}{9}\times\frac{4}{8}=\frac{1}{6}\)
   • \(P(X=1)=3 \times \frac{4}{10}\times\frac{6}{9}\times\frac{5}{8}=\frac{1}{2}\)
   • \(P(X=2)=3 \times \frac{4}{10}\times\frac{3}{9}\times\frac{6}{8}=\frac{3}{10}\)
   • \(P(X=3)=\frac{4}{10}\times\frac{3}{9}\times\frac{2}{8}=\frac{1}{30}\)

2. (b) Find E(X).

   • \(E(X)=0\times\frac{1}{6}+1\times\frac{1}{2}+2\times\frac{3}{10}+3\times\frac{1}{30}=\frac{6}{5}\)
   • Check: 3 × 4/10 = 6/5.

The table shows the probability distribution of Y, the number of rooms booked at a small hotel in an evening.

1. (a) Find the value of k.

   • \(k+0.2+2k+0.25+0.1=1\)
   • \(3k=0.45\)
   • \(k=0.15\)

2. (b) Find E(Y).

   • \(E(Y)=1\times0.15 + 2\times0.2 + 3\times0.3 + 4\times0.25 + 5\times0.1=2.95\)

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

   • E(Y) = 2.95, so Y > 2.95 means y = 3, 4, 5.
   • \(P(Y>E(Y))=0.3+0.25+0.1=0.65\)