Fair games and further distributions Problem set B One question at a time

You pay £2.50 to play a game. A fair four-sided die (faces 1 to 4) is rolled and you receive the score in pounds.

1. (a) Find the expected payout.

   • E(payout) = 1×1/4 + 2×1/4 + 3×1/4 + 4×1/4 = £2.50

2. (b) Is the game fair? Give a reason.

   • Expected gain = £2.50 − £2.50 = £0.00
   • A fair game has expected gain 0.

A discrete random variable X has \(P(X=x)=\frac{x^2+2}{22}\) for x = 0, 1, 2, 3.

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

   • \(P(X=0)=\frac{0+2}{22}=\frac{1}{11}\)
   • \(P(X=1)=\frac{1+2}{22}=\frac{3}{22}\)
   • \(P(X=2)=\frac{4+2}{22}=\frac{3}{11}\)
   • \(P(X=3)=\frac{9+2}{22}=\frac{1}{2}\)
   • The probabilities add to 1.

2. (b) Find E(X).

   • \(E(X)=0\times\frac{1}{11}+1\times\frac{3}{22}+2\times\frac{3}{11}+3\times\frac{1}{2}=\frac{24}{11}=2.18\)

A fair four-sided die (faces 1 to 4) is rolled and a fair five-sided spinner (sectors 1 to 5) is spun. X is the sum of the two scores. The table shows every outcome.

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

   • There are 20 equally likely outcomes. Count how many give each total (see the table).
   • Outcomes for each total: 2: 1, 3: 2, 4: 3, 5: 4, 6: 4, 7: 3, 8: 2, 9: 1. Each is out of 20.

2. (b) Find the expected total score.

   • \(E(X)=2\times\frac{1}{20}+3\times\frac{1}{10}+4\times\frac{3}{20}+5\times\frac{1}{5}+6\times\frac{1}{5}+7\times\frac{3}{20}+8\times\frac{1}{10}+9\times\frac{1}{20}=5.5\)

A basketball player scores (S) a free throw with probability 0.7. She takes three throws. X is the number of throws scored. The tree diagram shows the outcomes.

1. (a) Find the probability of exactly two throws scored.

   • Three routes give two: SSM, SMS, MSS.
   • \(3\times0.7^2\times0.3=0.441\)

2. (b) Find the probability distribution of X.

   • \(P(X=0)=0.3^3=0.027\)
   • \(P(X=1)=3\times0.7\times0.3^2=0.189\)
   • \(P(X=2)=0.441\)
   • \(P(X=3)=0.7^3=0.343\)

3. (c) Find E(X).

   • \(E(X)=0\times0.027+1\times0.189+2\times0.441+3\times0.343=2.1\)
   • Check: 3 × 0.7 = 2.1.

Two names are drawn from a hat of 10 names without replacement. 3 of the names are Year 12 students. X is the number of Year 12 names drawn.

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

   • \(P(X=0)=\frac{7}{10}\times\frac{6}{9}=\frac{7}{15}\)
   • \(P(X=1)=2\times\frac{3}{10}\times\frac{7}{9}=\frac{7}{15}\)
   • \(P(X=2)=\frac{3}{10}\times\frac{2}{9}=\frac{1}{15}\)

2. (b) Find E(X).

   • \(E(X)=0+1\times\frac{7}{15}+2\times\frac{1}{15}=\frac{3}{5}=0.6\)

At a fair, two spinners are each numbered 1 to 5 and spun together. The score X is the positive difference between the two numbers. The table shows every outcome.

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

   • There are 25 equally likely outcomes. Count how many cells show each difference (see the table).
   • Outcomes for each value: 0: 5, 1: 8, 2: 6, 3: 4, 4: 2. Each is out of 25.

2. (b) Find the expected score E(X).

   • \(E(X)=0\times\frac{1}{5}+1\times\frac{8}{25}+2\times\frac{6}{25}+3\times\frac{4}{25}+4\times\frac{2}{25}=1.6\)

At a summer fair, 100 raffle tickets are sold at $1 each. There is one prize of $40 and two prizes of $10.

1. (a) Find the expected prize for one ticket.

   • Total prizes = 1×40 + 2×10 = 60
   • \(E(\text{prize})=\frac{60}{100}=0.60\)

2. (b) Find the expected gain for a ticket buyer. Is the raffle fair?

   • Expected gain = $0.60 − $1.00 = −$0.40
   • Not 0, so not fair: it favours the organiser.

3. (c) Find the organiser's total profit if all the tickets are sold.

   • Takings = 100 × $1.00 = $100.00
   • Profit = $100.00 − $60 = $40.00

A player rolls a fair six-sided die. If it shows a 6 the player wins $15. Otherwise the player pays $n. Find the value of n that makes the game fair.

• \(P(\text{win})=\frac{1}{6},\quad P(\text{lose})=\frac{5}{6}\)
• \(\frac{1}{6}\times15-\frac{5}{6}\times n=0\)
• \(15=5n\Rightarrow n=3.00\)

The number of absent students, X, in a class on a given day has \(P(X=x)=k(6-x)\) for x = 0, 1, 2, 3, 4, where k is a constant.

1. (a) Find the value of k.

   • \(k(6)+k(5)+k(4)+k(3)+k(2)=20k=1\)
   • \(k=\frac{1}{20}\)

2. (b) Find P(X ≤ 1).

   • \(P(X\le1)=\frac{3}{10}+\frac{1}{4}=\frac{11}{20}\)

3. (c) Find E(X).

   • \(E(X)=0\times\frac{3}{10}+1\times\frac{1}{4}+2\times\frac{1}{5}+3\times\frac{3}{20}+4\times\frac{1}{10}=\frac{3}{2}=1.5\)

The table shows the probability distribution of X. The expected value of X is 1.7.

1. (a) Write down two equations in a and b.

   • The probabilities add to 1: a + b = 0.55
   • E(X) = 1.7: 0×0.2 + 1×a + 2×b + 3×0.25 = 1.7
   • a + 2b = 0.95 (after collecting the known terms)

2. (b) Find the values of a and b.

   • Solve the pair on the calculator, or by substitution.
   • a = 0.15, b = 0.4
   • Both are between 0 and 1, so they are valid.

A discrete random variable X has \(P(X=x)=cx^2\) for x = 2, 3, 4, 5, where c is a constant.

1. (a) Find the value of c.

   • \(c(2^2+3^2+4^2+5^2)=54c=1\)
   • \(c=\frac{1}{54}\)

2. (b) Find P(X ≥ 4 | X ≥ 3).

   • X ≥ 3 leaves x = 3, 4, 5.
   • \(P(X\ge4\mid X\ge3)=\frac{\frac{41}{54}}{\frac{25}{27}}=\frac{41}{50}=0.82\)

3. (c) Find E(X).

   • \(E(X)=2\times\frac{2}{27}+3\times\frac{1}{6}+4\times\frac{8}{27}+5\times\frac{25}{54}=\frac{112}{27}=4.15\)