Probability Distributions

Have a go first. Then open the working one line at a time.

1. Expected payout

   \(E(\text{win})=20\times\frac{1}{8}+6\times\frac{3}{8}+0\times\frac{1}{2}=4.75\)

   Each prize times its probability.

2. Expected gain

   \(\text{gain}=4.75-5=-0.25\)

   Subtract the stake. A fair game has expected gain 0.

3. Conclude

   Expected gain = −0.25 dollars, not 0

   Not fair. On average the player loses 25 cents a game, so it favours the organiser.

Have a go first. Then open the working one line at a time.

1. Probabilities

   \(P(\text{picture card})=\frac{12}{52},\quad P(\text{other})=\frac{40}{52}\)

   There are 3 × 4 = 12 picture cards.

2. Expected gain = 0

   \(\frac{12}{52}\times9-\frac{40}{52}\times n=0\)

   A loss of $n counts as −n.

3. Solve

   \(108=40n\Rightarrow n=2.7\)

   Multiply through by 52, then divide by 40. The player pays $2.70 on losing.

Have a go first. Then open the working one line at a time.

1. Probabilities

   \(P(0)=\frac{40}{52}\times\frac{39}{51}=\frac{10}{17},\ P(1)=2\times\frac{12}{52}\times\frac{40}{51}=\frac{80}{221},\ P(2)=\frac{12}{52}\times\frac{11}{51}=\frac{11}{221}\)

   The second card is out of 51.

2. Table

   \(\frac{10}{17}+\frac{80}{221}+\frac{11}{221}=1\)

   The total is 1.

3. E(X)

   \(E(X)=0\times\frac{10}{17}+1\times\frac{80}{221}+2\times\frac{11}{221}=\frac{6}{13}=0.462\)

   Check: 2 cards, and 12 of 52 are picture cards: 2 × 12/52.

Have a go first. Then open the working one line at a time.

1. All the outcomes

   25 equally likely outcomes. Each cell shows the larger score.

   Rows are the first spinner, columns the second.

2. Count

   Number of cells for each value: Y = 1: 1, Y = 2: 3, Y = 3: 5, Y = 4: 7, Y = 5: 9, each out of 25.

   The larger score is more likely to be a high value, so the table is not even.

3. E(Y)

   \(E(Y)=1\times\frac{1}{25}+2\times\frac{3}{25}+3\times\frac{1}{5}+4\times\frac{7}{25}+5\times\frac{9}{25}=\frac{19}{5}=3.8\)

   Check: more than the mean of one spinner, which is 3.

Have a go first. Then open the working one line at a time.

1. Values and k

   \(k(1)+k(3)+k(5)+k(7)=16k=1\Rightarrow k=\frac{1}{16}\)

   Put x = 0, 1, 2, 3 into 2x + 1, add, and set the total to 1.

2. Table

   P(X = x) = 1/16, 3/16, 5/16, 7/16

   Evaluate the formula at each x.

3. P(X ≥ 2)

   \(\frac{5}{16}+\frac{7}{16}=\frac{3}{4}\)

   Add the last two.

4. Conditional

   \(P(X=3\mid X\ge1)=\frac{\frac{7}{16}}{\frac{15}{16}}=\frac{7}{15}=0.467\)

   X ≥ 1 leaves x = 1, 2, 3. Their total, 15/16, is the denominator.

5. E(X)

   \(E(X)=0\times\frac{1}{16}+1\times\frac{3}{16}+2\times\frac{5}{16}+3\times\frac{7}{16}=\frac{17}{8}=2.125\)

   Multiply each x by its probability and add.

Have a go first. Then open the working one line at a time.

1. Sum to 1

   \(0.1+a+0.35+b=1\Rightarrow a+b=0.55\)

   First equation.

2. Use E(X)

   \(1(0.1)+2a+3(0.35)+4b=2.85\Rightarrow 2a+4b=1.7\)

   Second equation. Multiply each x by its probability and set the total equal to 2.85.

3. Solve the pair

   a + b = 0.55 and 2a + 4b = 1.7: a = 0.25, b = 0.3

   Solve with the equation solver on the calculator, or by substitution. Both values are between 0 and 1.