Discrete random variables and distribution tables Problem set B Worked solutions

A drawing pin lands point up with probability 0.35. It is dropped twice. X is the number of times it lands point up.

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

   • \(P(X=0)=0.65\times0.65=0.4225\)
   • \(P(X=1)=2\times0.35\times0.65=0.455\)
   • \(P(X=2)=0.35\times0.35=0.1225\)
   • The probabilities add to 1.

2. (b) Find P(X ≥ 1).

   • \(P(X\ge1)=1-P(X=0)=1-0.4225=0.5775\)

A wheel is split into 12 equal sectors numbered 1 to 12. It is spun once and X is the number it stops on.

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

   • The 12 scores are equally likely.
   • \(P(X=x)=\frac1{12}\)

2. (b) Find P(X ≥ 10).

   • 3 scores are 10 or more.
   • \(P(X\ge10)=\frac{3}{12}=\frac{1}{4}\)

3. (c) Find P(4 ≤ X ≤ 7).

   • 4 scores from 4 to 7.
   • \(P(4\le X\le7)=\frac{4}{12}=\frac{1}{3}\)

The table shows the probability distribution of X, the number of siblings of a student in a class.

1. (a) Find the value of k.

   • \(0.22 + 0.38 + 0.09 + 0.05+k=1\)
   • \(k=1-0.74=0.26\)

2. (b) Find P(X ≤ 1).

   • \(P(X\le1)=P(X=0)+P(X=1)=0.22+0.38=0.6\)

3. (c) Find P(X ≥ 3).

   • \(P(X\ge3)=P(X=3)+P(X=4)=0.09+0.05=0.14\)

The probability that a train is late on any one day is 0.2. The days are independent. X is the number of late trains on a three-day trip.

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

   • \(P(X=0)=0.8^3=0.512\)
   • \(P(X=1)=3\times0.2\times0.8^2=0.384\)
   • \(P(X=2)=3\times0.2^2\times0.8=0.096\)
   • \(P(X=3)=0.2^3=0.008\)
   • The four probabilities add to 1.

2. (b) Find P(X ≥ 2).

   • \(P(X\ge2)=P(X=2)+P(X=3)=0.096+0.008=0.104\)

A teacher chooses 2 students at random from a group of 3 boys and 6 girls. X is the number of boys chosen.

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

   • \(P(X=0)=\frac{6}{9}\times\frac{5}{8}=\frac{5}{12}\)
   • \(P(X=1)=2\times\frac{3}{9}\times\frac{6}{8}=\frac{1}{2}\)
   • \(P(X=2)=\frac{3}{9}\times\frac{2}{8}=\frac{1}{12}\)
   • Check: the three probabilities add to 1.

2. (b) Find P(X ≥ 1).

   • \(P(X\ge1)=1-P(X=0)=1-\frac{5}{12}=\frac{7}{12}\)

Three cards are drawn one at a time, with replacement, from ten cards numbered 1 to 10. X is the number of times the card showing 10 is drawn.

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

   • \(P(X=0)=\left(0.9\right)^3=0.729\)
   • \(P(X=1)=3\times0.1\times\left(0.9\right)^2=0.243\)
   • \(P(X=2)=3\times\left(0.1\right)^2\times0.9=0.027\)
   • \(P(X=3)=\left(0.1\right)^3=0.001\)
   • The four probabilities add to 1.

2. (b) Find P(X ≥ 1).

   • \(P(X\ge1)=1-P(X=0)=0.271\)

The table shows the probability distribution of R, the number of puncture repairs a cycle shop does in a day.

1. (a) Find the value of k.

   • \(0.21 + 0.33 + 0.24 + 0.13+k=1\)
   • \(k=0.09\)

2. (b) Find P(R ≥ 2).

   • \(P(R\ge2)=0.24+0.13+0.09=0.46\)

3. (c) Find P(R ≥ 3 | R ≥ 2).

   • \(P(R\ge3\mid R\ge2)=\frac{P(R\ge3)}{P(R\ge2)}=\frac{0.22}{0.46}=0.478\)

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

1. (a) Show that k² + 3k − 0.31 = 0.

   • \(k^2+k+2k+0.69=1\)
   • \(k^2+3k-0.31=0\)

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

   • Polynomial solver, coefficients 1, 3, −0.31: k = 0.1 or k = −3.1
   • A probability cannot be negative, so reject the negative root.
   • \(k=0.1\)

3. (c) Find P(X ≥ 3).

   • \(P(X\ge3)=P(X=3)+P(X=4)=2k+0.69=0.89\)

A team of 3 is chosen at random from 5 girls and 3 boys. X is the number of boys in the team.

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

   • \(P(X=0)=\frac{5}{8}\times\frac{4}{7}\times\frac{3}{6}=\frac{5}{28}\)
   • \(P(X=1)=3 \times \frac{3}{8}\times\frac{5}{7}\times\frac{4}{6}=\frac{15}{28}\)
   • \(P(X=2)=3 \times \frac{3}{8}\times\frac{2}{7}\times\frac{5}{6}=\frac{15}{56}\)
   • \(P(X=3)=\frac{3}{8}\times\frac{2}{7}\times\frac{1}{6}=\frac{1}{56}\)
   • Check: the four probabilities add to 1.

2. (b) Find P(X ≥ 1).

   • \(P(X\ge1)=1-P(X=0)=1-\frac{5}{28}=\frac{23}{28}\)

3. (c) Find P(X = 2 | X ≥ 1).

   • \(P(X=2\mid X\ge1)=\frac{\frac{15}{56}}{\frac{23}{28}}=\frac{15}{46}=0.326\)