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

You pay $1.50 to play a game. Three fair coins are tossed and you receive $1 for each head.

1. (a) Find the expected payout.

   • E(payout) = 0×1/8 + 1×3/8 + 2×3/8 + 3×1/8 = $1.50

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

   • Expected gain = $1.50 − $1.50 = $0.00
   • A fair game has expected gain 0.

A discrete random variable X has \(P(X=x)=\frac{x^2+1}{18}\) for x = 0, 1, 2, 3.

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

   • \(P(X=0)=\frac{0+1}{18}=\frac{1}{18}\)
   • \(P(X=1)=\frac{1+1}{18}=\frac{1}{9}\)
   • \(P(X=2)=\frac{4+1}{18}=\frac{5}{18}\)
   • \(P(X=3)=\frac{9+1}{18}=\frac{5}{9}\)
   • The probabilities add to 1.

2. (b) Find E(X).

   • \(E(X)=0\times\frac{1}{18}+1\times\frac{1}{9}+2\times\frac{5}{18}+3\times\frac{5}{9}=\frac{7}{3}=2.33\)

A fair four-sided die (faces 1 to 4) is rolled and a fair three-sided spinner (sectors 1 to 3) 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 12 equally likely outcomes. Count how many give each total (see the table).
   • Outcomes for each total: 2: 1, 3: 2, 4: 3, 5: 3, 6: 2, 7: 1. Each is out of 12.

2. (b) Find the expected total score.

   • \(E(X)=2\times\frac{1}{12}+3\times\frac{1}{6}+4\times\frac{1}{4}+5\times\frac{1}{4}+6\times\frac{1}{6}+7\times\frac{1}{12}=4.5\)

A weighted coin lands heads (H) with probability 0.4. It is tossed three times. X is the number of heads. The tree diagram shows the outcomes.

1. (a) Find the probability of exactly two heads.

   • Three routes give two: HHT, HTH, THH.
   • \(3\times0.4^2\times0.6=0.288\)

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

   • \(P(X=0)=0.6^3=0.216\)
   • \(P(X=1)=3\times0.4\times0.6^2=0.432\)
   • \(P(X=2)=0.288\)
   • \(P(X=3)=0.4^3=0.064\)

3. (c) Find E(X).

   • \(E(X)=0\times0.216+1\times0.432+2\times0.288+3\times0.064=1.2\)
   • Check: 3 × 0.4 = 1.2.

Two cards are dealt from a standard pack of 52 without replacement. X is the number of aces dealt.

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

   • \(P(X=0)=\frac{48}{52}\times\frac{47}{51}=\frac{188}{221}\)
   • \(P(X=1)=2\times\frac{4}{52}\times\frac{48}{51}=\frac{32}{221}\)
   • \(P(X=2)=\frac{4}{52}\times\frac{3}{51}=\frac{1}{221}\)

2. (b) Find E(X).

   • \(E(X)=0+1\times\frac{32}{221}+2\times\frac{1}{221}=\frac{2}{13}=0.154\)

In a board game, two fair four-sided dice (faces 1 to 4) are rolled. 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 16 equally likely outcomes. Count how many cells show each difference (see the table).
   • Outcomes for each value: 0: 4, 1: 6, 2: 4, 3: 2. Each is out of 16.

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

   • \(E(X)=0\times\frac{1}{4}+1\times\frac{3}{8}+2\times\frac{1}{4}+3\times\frac{1}{8}=1.25\)

A school raffle sells 200 tickets at $2 each. There is one prize of $150 and three prizes of $20.

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

   • Total prizes = 1×150 + 3×20 = 210
   • \(E(\text{prize})=\frac{210}{200}=1.05\)

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

   • Expected gain = $1.05 − $2.00 = −$0.95
   • Not 0, so not fair: it favours the organiser.

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

   • Takings = 200 × $2.00 = $400.00
   • Profit = $400.00 − $210 = $190.00

A player draws one card at random from a standard pack. If it is a spade the player wins $6. Otherwise the player pays $n. Find the value of n that makes the game fair.

• \(P(\text{win})=\frac{1}{4},\quad P(\text{lose})=\frac{3}{4}\)
• \(\frac{1}{4}\times6-\frac{3}{4}\times n=0\)
• \(78=39n\Rightarrow n=2.00\)

The number of times, X, a printer jams in a day has \(P(X=x)=k(5-x)\) for x = 0, 1, 2, 3, 4, where k is a constant.

1. (a) Find the value of k.

   • \(k(5)+k(4)+k(3)+k(2)+k(1)=15k=1\)
   • \(k=\frac{1}{15}\)

2. (b) Find P(X ≤ 1).

   • \(P(X\le1)=\frac{1}{3}+\frac{4}{15}=\frac{3}{5}\)

3. (c) Find E(X).

   • \(E(X)=0\times\frac{1}{3}+1\times\frac{4}{15}+2\times\frac{1}{5}+3\times\frac{2}{15}+4\times\frac{1}{15}=\frac{4}{3}=1.33\)

The table shows the probability distribution of X. The expected value of X is 2.45.

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

   • The probabilities add to 1: a + b = 0.65
   • E(X) = 2.45: 1×a + 2×0.25 + 3×b + 4×0.1 = 2.45
   • a + 3b = 1.55 (after collecting the known terms)

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

   • Solve the pair on the calculator, or by substitution.
   • a = 0.2, b = 0.45
   • Both are between 0 and 1, so they are valid.

A discrete random variable X has \(P(X=x)=cx^2\) for x = 1, 2, 3, 4, where c is a constant.

1. (a) Find the value of c.

   • \(c(1^2+2^2+3^2+4^2)=30c=1\)
   • \(c=\frac{1}{30}\)

2. (b) Find P(X ≥ 3 | X ≥ 2).

   • X ≥ 2 leaves x = 2, 3, 4.
   • \(P(X\ge3\mid X\ge2)=\frac{\frac{5}{6}}{\frac{29}{30}}=\frac{25}{29}=0.862\)

3. (c) Find E(X).

   • \(E(X)=1\times\frac{1}{30}+2\times\frac{2}{15}+3\times\frac{3}{10}+4\times\frac{8}{15}=\frac{10}{3}=3.33\)