Independent events Problem set B Worked solutions

On a school day, the probability that the lift is out of order is 0.55 and the probability that the canteen card machine is out of order is 0.3. The two faults are independent. Find the probability that both are out of order on the same day.

• Independent, so \(P(A\cap B)=P(A)\times P(B)\)
• \(0.55\times 0.3=0.165\)

For events A and B, \(P(A)=0.3\), \(P(B)=0.6\) and \(P(A\cup B)=0.75\). Determine whether A and B are independent. Justify your answer.

• \(P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.3+0.6-0.75=0.15\)
• \(P(A)\times P(B)=0.3\times 0.6=0.18\)
• The two are not equal, so the events are not independent.

A bike's front light and its rear light fail independently. The probability that the front light fails is 0.2. The probability that both fail is 0.09. Find the probability that the rear light fails.

• Independent, so \(P(A\cap B)=P(A)\times P(B)\)
• \(0.09=0.2\times P(B)\)
• \(P(B)=0.09\div 0.2=0.45\)

Events A and B are independent, with \(P(A)=\frac{5}{9}\) and \(P(A\cap B)=\frac{10}{27}\). Find \(P(B)\).

• Independent, so \(P(A\cap B)=P(A)\times P(B)\)
• \(\frac{10}{27}=\frac{5}{9}\times P(B)\)
• \(P(B)=\frac{10}{27}\div \frac{5}{9}=\frac{2}{3}\)

Events C and D have P(C) = 0.4 and P(D) = 0.3.

1. (a) C and D are independent. Find P(C ∪ D).

   • Independent: \(P(C\cap D)=0.4\times0.3=0.12\)
   • \(P(C\cup D)=0.4+0.3-0.12=0.58\)

2. (b) State, with a reason, whether C and D could also be mutually exclusive.

   • Mutually exclusive needs \(P(C\cap D)=0\).
   • But \(P(C\cap D)=0.12\), so they cannot be both.

At a school fair, two fair six-sided dice are rolled and the prize depends on the product of the two numbers.

1. (a) Use a sample space diagram to find the probability that the score is at most 6.

   • Draw a 6 by 6 grid with the first die down the side and the second across the top. Each cell holds the product.
   • 14 of the 36 products are at most 6.
   • \(P=\frac{14}{36}=\frac{7}{18}\)

2. (b) The game is played 108 times. How many times would you expect the score to be at most 6?

   • Expected number = probability × number of trials
   • \(\frac{7}{18}\times 108=42\)

A game uses a fair four-sided die (numbered 1 to 4) and a fair eight-sided die (numbered 1 to 8). The two scores are added to give the total.

1. (a) Find the probability that the total is 8.

   • Grid of totals: 4 rows (the 4-sided die) and 8 columns (the 8-sided die), 32 outcomes in all.
   • 4 of them fit.
   • \(P=\frac{4}{32}=\frac{1}{8}\)

2. (b) Find the probability that the total is a multiple of 3.

   • Grid of totals: 4 rows (the 4-sided die) and 8 columns (the 8-sided die), 32 outcomes in all.
   • 11 of them fit.
   • \(P=\frac{11}{32}\)

Of 160 residents of a street, 70 own a car, 55 own a bicycle and 25 own both.

1. (a) Draw a Venn diagram to show this information.

   • Start in the middle: 25 do both.
   • Only a car: 70 − 25 = 45. Only a bicycle: 55 − 25 = 30.
   • Neither: 160 − (45 + 25 + 30) = 60

2. (b) Find the probability that a person chosen at random does neither.

   • \(P=\frac{60}{160}=\frac{3}{8}\)

3. (c) Find the probability that a resident owns a car, given that they do not own a bicycle.

   • Not owning a bicycle means outside the bicycle circle: 45 + 60 = 105 people.
   • \(P=\frac{45}{105}=\frac{3}{7}\)

A fair 12-sided die is numbered 1 to 12 and rolled once. M is the event "the score is a multiple of 3" and E is the event "the score is even".

1. (a) Are M and E mutually exclusive? Give a reason.

   • M = {3, 6, 9, 12} and E = {2, 4, 6, 8, 10, 12}. The scores 6 and 12 are in both, so \(P(M\cap E)=\frac{2}{12}\neq0\).

2. (b) Are M and E independent? Show your working.

   • \(P(M)=\frac4{12}=\frac13\), \(P(E)=\frac6{12}=\frac12\)
   • \(P(M)\times P(E)=\frac16=\frac2{12}=P(M\cap E)\), so M and E are independent.

Two events are independent, with \(P(A)=0.4\) and \(P(B)=0.25\). Find \(P(A\cup B)\).

• Independent, so \(P(A\cap B)=0.4\times 0.25=0.1\)
• \(P(A\cup B)=0.4+0.25-0.1=0.55\)

Events A and B are independent, with \(P(A)=0.7\) and \(P(A\cup B)=0.91\). Find \(P(B)\).

• Independent: \(P(A\cap B)=P(A)P(B)\)
• \(P(A\cup B)=P(A)+P(B)-P(A)P(B)\), so \(0.91=0.7+P(B)(1-0.7)\)
• \(P(B)=\frac{0.91-0.7}{0.3}=0.7\)