# Exam-style Questions: Probability Distributions

**1.** Two fair dice are thrown. If the scores are unequal, the larger of the two scores is recorded. If the scores are equal then that score is recorded. Let X denote the number recorded.

**a)** show that P(X = 2) = 1/12 and draw up a table showing the probability distribution of X

**(4 marks)**

**b)** find the mean and variance of

**(4 marks)**

**(Marks available: 8)**

**Answer outline and marking scheme for question: 1**

**a)** P(X = 2) = P( 1 and 2) or P(2 and 1) or P(2 and 2)

= (1/6 x 1/6) + (1/6 x 1/6) + (1/6 x 1/6) = 3/36 A1

= 1/12 as required A1

x |
1 | 2 | 3 | 4 | 5 | 6 |
---|---|---|---|---|---|---|

P(X=x) |
1/36 | 1/12 | 5/36 | 7/36 | 9/36 | 11/36 |

**(4 marks)**

**b)** E(X) = (1 x 1/36) + (2 x 1/12) + (3 x 5/36) + (4 x 7/36) + (5 x 9/36) + (6 x 11/36)

= 161/36

= 4.47 (2dp)

for Var(X) we first need

E(X2) = (12 x 1/36) + (22 x 1/12) + (32 x 5/36) + (42 x 7/36) + (52 x 9/36) + (62 x 11/36)

= 791/36 = 21.97 (2dp)

Var(X) = E(X2) - E(X)2

= 21.97 - 4.472

= 1.97

**(4 marks)**

**(Marks available: 8)**

**2.** It is known that 65% of sixth formers think their maths teacher is cool. In a certain sample, 10 pupils are picked at random and it is required to calculate the probability that less than 3 of the 10 think that their teacher is cool. Name a probability distribution that can be used for modelling this situation stating one necessary assumption for this model to be valid.

**a)** use the model to calculate the required probability

**(4 marks)**

**b)** calculate the mean and variance of this distribution

**(3 marks)**

**(Marks available: 7)**

**Answer outline and marking scheme for question: 2**

Binomial distribution with n = 10 and probability of success, p = 0.65

assume each pupil picked is independent of each other

**a)** using X ~ B(10, 0.65)

need P(X

=0.3510 + 10(0.35)9(0.65)^{10}C_{2} + (0.35)8(0.65)2

= 0.0048 (2sf)

**(4 marks)**

**b)** E(X) = n x p = 10 x 0.65 = 6.5 A1

Var(X) = n x p x q = 10 x 0.65 x 0.35 = 2.275

**(3 marks)**

**(Marks available: 7)**

**3.** When I ask a sixth form class a question the probability that I get an answer is 0.36. if I don't get an answer first time I keep trying until I am blessed with an answer. Let X denote the number of attempts I have to make in order to get an answer. Stating any assumption, identify the probability distribution of X.

**(2 marks)**

**hence calculate:**

**a)** P(X = 5)

**(2 marks)**

**b)** P(X = 4)

**(3 marks)**

**c)** E(X) and Var(X)

**(3 marks)**

**(Marks available: 10)**

**Answer outline and marking scheme for question: 3**

**3.** Geometric distribution with probability of success, p = 0.36

assume independence of sixth formers asked

**(2 marks)**

**a)** P(X = 5) = (0.64)^{4} x 0.36 = 0.0604 (3sf)

**(2 marks)**

**b)** X ~ Geo(0.36)

P(X = 4) = P(X > 5)

= q^{5} = 0.64^{5}

= 0.107 (3sf)

**(3 marks)**

**c)** E(X) = 1/p = 1/0.36 = 2.78 (3sf)

Var(X) = q/p^{2}

= 0.64/0.362 = 4.94

**(3 marks)**

**(Marks available: 10)**