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7.17 Conditional Probability

EDEXCEL 1MA1 · Calculator allowed · about 100 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

The probability that event A occurs given that event B has already occurred is written P(A|B).

(a)Which of these gives the correct formula for P(A|B)?(1)

A) P(A) x P(B)
B) P(A and B) / P(B)
C) P(A and B) / P(A)
D) P(A) / P(B)

(Total for Question 1 is 1 mark)

150 members of a gym were asked whether they prefer cardio or weights training. The two-way table shows the results.

Cardio Weights Total
Under 30 38 22 60
30 or over 34 56 90
Total 72 78 150

(a)One member is chosen at random from those aged 30 or over. Find the probability that this member prefers weights training.(2)

(b)One member is chosen at random from those who prefer cardio. Find the probability that this member is aged under 30.(2)

(Total for Question 2 is 4 marks)

100 students at a school study either French or Spanish. The two-way table shows some information about the students, split by gender.

French Spanish Total
Boys 21 ? 48
Girls 25 ? 52
Total ? ? 100

(a)Complete the two-way table.(2)

(b)One student is chosen at random from the girls. Find the probability that this student studies Spanish.(2)

(Total for Question 3 is 4 marks)

A fair coin is flipped and, independently, a fair six-sided dice is rolled. Event A is 'the coin shows Heads'. Event B is 'the dice shows a number greater than 4'.

(a)Find P(A), P(B) and P(A and B).(3)

(b)Show that P(A|B) = P(A), and state what this tells you about events A and B.(3)

(Total for Question 4 is 6 marks)

In a survey of 60 students, G is the event 'plays football' and T is the event 'plays tennis'. The Venn diagram shows n(G) = 32, n(T) = 25, n(G and T) = 14, and n(neither) = 17.

(a)One student is picked at random. Find P(T).(1)

(b)Given that the student plays football, find the probability that they also play tennis.(2)

(c)Given that the student does not play tennis, find the probability that they play football.(2)

(Total for Question 5 is 5 marks)

A class committee of 10 students contains 6 girls and 4 boys. Two students are chosen at random from the committee, one after another, without replacement.

(a)Find the probability that both students chosen are girls.(2)

(b)Find the probability that at least one boy is chosen.(3)

(Total for Question 6 is 5 marks)

A bag contains 9 counters: 5 red and 4 blue. Two counters are taken at random from the bag, one after another, without replacement.

(a)Complete the tree diagram by writing down the four second-pick probabilities.(2)

(b)Find the probability that both counters are the same colour.(3)

(c)Given that both counters are the same colour, find the probability that both are red.(3)

(Total for Question 7 is 8 marks)

A bag contains 10 cards, numbered 1 to 10. Two cards are drawn at random from the bag, one after another, without replacement.

(a)Given that the first card drawn shows an even number, find the probability that the second card drawn also shows an even number.(3)

(b)Find the probability that both cards drawn show even numbers.(2)

(Total for Question 8 is 5 marks)

For two events A and B, P(A) = 0.6, P(B) = 0.45 and P(A and B) = 0.3.

(a)Find P(A|B).(2)

(b)Find P(B|A).(2)

(c)Determine, showing your working, whether events A and B are independent.(2)

(Total for Question 9 is 6 marks)

A factory has two machines, A and B, producing circuit boards in one day. Machine A produces 120 boards, of which 6 are faulty. Machine B produces 80 boards, of which 12 are faulty.

(a)Complete a frequency tree to show this information, including the total number of faulty and non-faulty boards.(3)

(b)A board is chosen at random from the day's production and found to be faulty. Find the probability that it was produced by Machine B.(3)

(Total for Question 10 is 6 marks)

On any given day, the probability of rain depends on whether it rained the previous day. P(rain tomorrow | rain today) = 0.7. P(rain tomorrow | no rain today) = 0.25. It rained on Monday.

(a)Draw a tree diagram for Tuesday and Wednesday, and find the probability that it rains on both Tuesday and Wednesday.(2)

(b)Find the probability that it rains on at least one of Tuesday or Wednesday.(3)

(Total for Question 11 is 5 marks)

In a school of 48 students, A is the event 'studies Art' and M is the event 'studies Music'. On the Venn diagram, n(A and M) = x, the number studying only Art is 2x, the number studying only Music is 3x, and the number studying neither is 12.

(a)Show that x = 6.(2)

(b)Find P(A), the probability that a student chosen at random studies Art.(2)

(c)Given that a student studies Art, find the probability that they also study Music.(2)

(Total for Question 12 is 6 marks)

A bag contains 8 counters, which are either green or red. Two counters are taken at random from the bag, one after another, without replacement. The probability that both counters are green is 3/14.

(a)Show that there are 4 green counters in the bag.(3)

(Total for Question 13 is 3 marks)

200 patients are tested for a medical condition. The two-way table shows the test result and whether each patient actually has the condition.

Has condition No condition Total
Tests positive 18 14 32
Tests negative 2 166 168
Total 20 180 200

(a)One patient is chosen at random. Find the probability that they test positive.(1)

(b)Given that a patient tests positive, find the probability that they actually have the condition.(2)

(c)Given that a patient does not have the condition, find the probability that they test positive.(2)

(Total for Question 14 is 5 marks)

A bag contains 10 counters, r of which are red; the rest are green. Two counters are taken at random from the bag, one after another, without replacement. The probability that both counters are red is 1/15.

(a)Find the value of r.(4)

(Total for Question 15 is 4 marks)

A factory buys components from three suppliers. Supplier X provides 50% of components, of which 2% are defective. Supplier Y provides 30%, of which 5% are defective. Supplier Z provides 20%, of which 8% are defective. A component is chosen at random and found to be defective.

(a)Find the probability that the component came from Supplier Z.(5)

(Total for Question 16 is 5 marks)

For any two events A and B, with P(B) not equal to 0 and P(A) not equal to 0.

(a)Prove that P(A|B) x P(B) = P(B|A) x P(A).(3)

(Total for Question 17 is 3 marks)

A drawer contains 6 blue socks and 4 black socks. Three socks are taken at random from the drawer, one after another, without replacement.

(a)Find the probability that all three socks taken are the same colour.(4)

(b)Given that all three socks taken are the same colour, find the probability that all three are blue.(2)

(Total for Question 18 is 6 marks)

Mark scheme · 7.17 Conditional Probability

Question 1

(a) B1 B
(a) Answer: B

Question 2

(a) M1 56/90 seen (oe unsimplified fraction)
(a) A1 28/45 oe cao
(a) Answer: 28/45
(b) M1 38/72 seen (oe unsimplified fraction)
(b) A1 19/36 oe cao
(b) Answer: 19/36

Question 3

(a) B1 Boys Spanish = 27 and Girls Spanish = 27
(a) B1 Total French = 46 and Total Spanish = 54
(a) Answer: Boys Spanish = 27, Girls Spanish = 27, Total French = 46, Total Spanish = 54
(b) M1 27/52 seen
(b) A1 27/52 cao (does not simplify)
(b) Answer: 27/52

Question 4

(a) B1 P(A) = 1/2
(a) B1 P(B) = 1/3 (numbers greater than 4 on a dice are 5 and 6)
(a) B1 P(A and B) = 1/6
(a) Answer: P(A) = 1/2, P(B) = 1/3, P(A and B) = 1/6
(b) M1 P(A|B) = P(A and B) / P(B) = (1/6) / (1/3)
(b) A1 = 1/2 = P(A) cso
(b) B1 states A and B are independent, because P(A|B) = P(A)
(b) Answer: P(A|B) = 1/2 = P(A), so A and B are independent.

Question 5

(a) B1 5/12 oe
(a) Answer: 5/12
(b) M1 14/32 seen
(b) A1 7/16 oe cao
(b) Answer: 7/16
(c) M1 n(not T) = 60 - 25 = 35, and only-football = 32 - 14 = 18 identified
(c) A1 18/35 oe cao
(c) Answer: 18/35

Question 6

(a) M1 6/10 x 5/9
(a) A1 1/3 oe cao
(a) Answer: 1/3
(b) M1 recognises at least one boy = 1 - P(both girls)
(b) M1 1 - 1/3
(b) A1 2/3 oe cao
(b) Answer: 2/3

Question 7

(a) B1 P(R|R) = 4/8 and P(B|R) = 4/8
(a) B1 P(R|B) = 5/8 and P(B|B) = 3/8
(a) Answer: After first red: P(R) = 4/8, P(B) = 4/8. After first blue: P(R) = 5/8, P(B) = 3/8.
(b) M1 5/9 x 4/8 (=20/72)
(b) M1 4/9 x 3/8 (=12/72)
(b) A1 4/9 oe cao (32/72 simplified)
(b) Answer: 4/9
(c) M1 numerator P(RR) = 20/72 identified
(c) M1 denominator P(same colour) = 32/72 from part (b) used
(c) A1 5/8 oe cao
(c) Answer: 5/8

Question 8

(a) M1 identifies 5 even numbers (2, 4, 6, 8, 10), so 4 even numbers remain
(a) M1 9 cards remain in total
(a) A1 4/9 cao
(a) Answer: 4/9
(b) M1 5/10 x 4/9
(b) A1 2/9 oe cao
(b) Answer: 2/9

Question 9

(a) M1 0.3 / 0.45
(a) A1 2/3 oe awrt 0.67
(a) Answer: 2/3 (awrt 0.67)
(b) M1 0.3 / 0.6
(b) A1 0.5 cao
(b) Answer: 0.5
(c) M1 P(A) x P(B) = 0.6 x 0.45 (= 0.27) calculated and compared with P(A and B) = 0.3
(c) A1 correct conclusion: not independent, since 0.27 does not equal 0.3 (oe using P(A|B) not equal to P(A))
(c) Answer: Not independent

Question 10

(a) B1 Machine A: 6 faulty, 114 non-faulty
(a) B1 Machine B: 12 faulty, 68 non-faulty
(a) B1 Totals: 18 faulty, 182 non-faulty, 200 boards overall
(a) Answer: A: 6 faulty, 114 non-faulty. B: 12 faulty, 68 non-faulty. Totals: 18 faulty, 182 non-faulty, 200 in total.
(b) M1 total faulty = 18 identified
(b) M1 12/18 formed
(b) A1 2/3 oe cao
(b) Answer: 2/3

Question 11

(a) M1 0.7 x 0.7
(a) A1 0.49 cao
(a) Answer: 0.49
(b) M1 P(no rain Tue) = 0.3 and P(no rain Wed | no rain Tue) = 0.75 identified
(b) M1 0.3 x 0.75 (= 0.225) and 1 - 0.225 used
(b) A1 0.775 cao
(b) Answer: 0.775

Question 12

(a) M1 2x + 3x + x + 12 = 48 formed (oe 6x + 12 = 48)
(a) A1 x = 6 cso
(a) Answer: x = 6
(b) M1 n(A) = 2x + x = 18 (using x = 6)
(b) A1 3/8 oe cao
(b) Answer: 3/8
(c) M1 6/18 seen
(c) A1 1/3 oe cao
(c) Answer: 1/3

Question 13

(a) M1 g/8 x (g-1)/7 = 3/14 formed, where g is the number of green counters
(a) M1 rearranges to g(g-1) = 12 (oe g² - g - 12 = 0)
(a) A1 solves to g = 4, rejecting g = -3, cso
(a) Answer: g = 4

Question 14

(a) B1 4/25 oe (= 0.16)
(a) Answer: 4/25
(b) M1 18/32 seen
(b) A1 9/16 oe cao
(b) Answer: 9/16
(c) M1 14/180 seen
(c) A1 7/90 oe cao
(c) Answer: 7/90

Question 15

(a) M1 r/10 x (r-1)/9 formed as the probability of both red
(a) M1 sets equal to 1/15 and rearranges to a 3-term quadratic equal to zero (r² - r - 6 = 0 oe)
(a) M1 attempts to solve the quadratic, e.g. factorises as (r-3)(r+2) = 0
(a) A1 r = 3 cao, rejecting r = -2 as r must be a positive whole number
(a) Answer: r = 3

Question 16

(a) M1 P(X and defective) = 0.5 x 0.02 = 0.01
(a) M1 P(Y and defective) = 0.3 x 0.05 = 0.015 and P(Z and defective) = 0.2 x 0.08 = 0.016
(a) M1 total P(defective) = 0.01 + 0.015 + 0.016 = 0.041
(a) M1 forms 0.016 / 0.041
(a) A1 16/41 oe awrt 0.390
(a) Answer: 16/41 (awrt 0.390)

Question 17

(a) M1 states P(A|B) = P(A and B)/P(B), so P(A|B) x P(B) = P(A and B)
(a) M1 states P(B|A) = P(A and B)/P(A), so P(B|A) x P(A) = P(A and B)
(a) A1 both expressions equal P(A and B), so P(A|B) x P(B) = P(B|A) x P(A), cso
(a) Answer: Proven: both sides equal P(A and B).

Question 18

(a) M1 P(BBB) = 6/10 x 5/9 x 4/8 (= 1/6)
(a) M1 P(KKK) = 4/10 x 3/9 x 2/8 (= 1/30)
(a) M1 adds the two probabilities: 1/6 + 1/30
(a) A1 1/5 oe cao
(a) Answer: 1/5
(b) M1 (1/6) / (1/5) formed
(b) A1 5/6 oe cao
(b) Answer: 5/6

Conditional Probability - Worksheets, Questions and Revision