Discrete random variables and distribution tables Problem set A Worked solutions

A biased coin lands heads with probability 0.7. It is tossed twice. X is the number of heads.

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

   • \(P(X=0)=0.3\times0.3=0.09\)
   • \(P(X=1)=2\times0.7\times0.3=0.42\)
   • \(P(X=2)=0.7\times0.7=0.49\)
   • The probabilities add to 1.

2. (b) Find P(X ≥ 1).

   • \(P(X\ge1)=1-P(X=0)=1-0.09=0.91\)

A fair 10-sided die has faces numbered 1 to 10. It is rolled once and X is the number shown.

1. (a) Write down P(X = x) for each score.

   • The 10 scores are equally likely.
   • \(P(X=x)=\frac1{10}\)

2. (b) Find P(X ≥ 8).

   • 3 scores are 8 or more.
   • \(P(X\ge8)=\frac{3}{10}\)

3. (c) Find P(3 ≤ X ≤ 6).

   • 4 scores from 3 to 6.
   • \(P(3\le X\le6)=\frac{4}{10}=\frac{2}{5}\)

The table shows the probability distribution of X, the number of cars owned by a household.

1. (a) Find the value of k.

   • \(0.12 + 0.41 + 0.15 + 0.04+k=1\)
   • \(k=1-0.72=0.28\)

2. (b) Find P(X ≤ 1).

   • \(P(X\le1)=P(X=0)+P(X=1)=0.12+0.41=0.53\)

3. (c) Find P(X ≥ 3).

   • \(P(X\ge3)=P(X=3)+P(X=4)=0.15+0.04=0.19\)

The probability that Sam cycles to school on any one day is 0.6. The days are independent. X is the number of days he cycles in a three-day school week.

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

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

2. (b) Find P(X ≥ 2).

   • \(P(X\ge2)=P(X=2)+P(X=3)=0.432+0.216=0.648\)

A bag holds 5 red counters and 4 white counters. Two counters are taken at random without replacement. X is the number of red counters taken.

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

   • \(P(X=0)=\frac{4}{9}\times\frac{3}{8}=\frac{1}{6}\)
   • \(P(X=1)=2\times\frac{5}{9}\times\frac{4}{8}=\frac{5}{9}\)
   • \(P(X=2)=\frac{5}{9}\times\frac{4}{8}=\frac{5}{18}\)
   • Check: the three probabilities add to 1.

2. (b) Find P(X ≥ 1).

   • \(P(X\ge1)=1-P(X=0)=1-\frac{1}{6}=\frac{5}{6}\)

A fair four-sided die has faces numbered 1 to 4. It is rolled three times and X is the number of times a 3 is rolled.

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

   • \(P(X=0)=\left(\frac{3}{4}\right)^3=\frac{27}{64}\)
   • \(P(X=1)=3\times\frac{1}{4}\times\left(\frac{3}{4}\right)^2=\frac{27}{64}\)
   • \(P(X=2)=3\times\left(\frac{1}{4}\right)^2\times\frac{3}{4}=\frac{9}{64}\)
   • \(P(X=3)=\left(\frac{1}{4}\right)^3=\frac{1}{64}\)
   • The four probabilities add to 1.

2. (b) Find P(X ≥ 1).

   • \(P(X\ge1)=1-P(X=0)=\frac{37}{64}\)

The table shows the probability distribution of G, the number of goals a team scores in a match.

1. (a) Find the value of k.

   • \(0.12 + 0.31 + 0.27 + 0.18+k=1\)
   • \(k=0.12\)

2. (b) Find P(G ≥ 2).

   • \(P(G\ge2)=0.27+0.18+0.12=0.57\)

3. (c) Find P(G ≥ 3 | G ≥ 2).

   • \(P(G\ge3\mid G\ge2)=\frac{P(G\ge3)}{P(G\ge2)}=\frac{0.3}{0.57}=0.526\)

A discrete random variable X has the distribution shown, where k is a constant.

1. (a) Show that k² + 3k − 0.64 = 0.

   • \(k^2+k+2k+0.36=1\)
   • \(k^2+3k-0.64=0\)

2. (b) Use your calculator to solve this equation and state the value of k.

   • Polynomial solver, coefficients 1, 3, −0.64: k = 0.2 or k = −3.2
   • A probability cannot be negative, so reject the negative root.
   • \(k=0.2\)

3. (c) Find P(X ≥ 3).

   • \(P(X\ge3)=P(X=3)+P(X=4)=2k+0.36=0.76\)

A box holds 4 red pencils and 3 blue pencils. Three pencils are taken at random without replacement. X is the number of red pencils taken.

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

   • \(P(X=0)=\frac{3}{7}\times\frac{2}{6}\times\frac{1}{5}=\frac{1}{35}\)
   • \(P(X=1)=3 \times \frac{4}{7}\times\frac{3}{6}\times\frac{2}{5}=\frac{12}{35}\)
   • \(P(X=2)=3 \times \frac{4}{7}\times\frac{3}{6}\times\frac{3}{5}=\frac{18}{35}\)
   • \(P(X=3)=\frac{4}{7}\times\frac{3}{6}\times\frac{2}{5}=\frac{4}{35}\)
   • Check: the four probabilities add to 1.

2. (b) Find P(X ≥ 1).

   • \(P(X\ge1)=1-P(X=0)=1-\frac{1}{35}=\frac{34}{35}\)

3. (c) Find P(X = 2 | X ≥ 1).

   • \(P(X=2\mid X\ge1)=\frac{\frac{18}{35}}{\frac{34}{35}}=\frac{9}{17}=0.529\)