A group of 12 people consists of 3 boys, 4 girls and 5 adults.

### (a) *Question*

In how many ways can a team of 5 people be chosen from the group if exactly one adult is included?

**Answer/Explanation**

Ans:

^{5}C_{1} ×^{ 7}C_{4}

175

### (b) *Question*

In how many ways can a team of 5 people be chosen from the group if the team includes at least 2 boys and at least 1 girl?

The same group of 12 people stand in a line.

**Answer/Explanation**

Ans:

2B 1G 2A ^{3}C_{2} × ^{4}C_{1} × ^{5}C_{2} = 120

2B 2G 1A ^{ 3}C_{2} × ^{4}C_{2} × ^{5}C_{1} = 90

2B 3G ^{3}C_{2} × ^{4}C_{3} = 12

3B 1G 1A ^{3}C_{3} × ^{4}C_{1} × ^{5}C_{1} = 20

3B 2G ^{3}C_{3} × ^{4}C_{2} = 6

[Total =] 248

### (c) *Question*

How many different arrangements are there in which the 3 boys stand together and an adult is at each end of the line?

**Answer/Explanation**

Ans:

8! × 3! ×^{ 5}P_{2}

4 838 400

**Question**

**Question**

The weights of apples of a certain variety are normally distributed with mean 82 grams. 22% of these apples have a weight greater than 87 grams.

(a) Find the standard deviation of the weights of these apples.

(b) Find the probability that the weight of a randomly chosen apple of this variety differs from the mean weight by less than 4 grams.

**Answer/Explanation**

**Ans:**

(a) \(P(X>87)=P(Z>\frac{87-82}{\sigma})=0.22\)

\(P(Z<\frac{5}{\sigma})=0.78\)

\((\frac{5}{\sigma}=)0.772\)

\(\sigma=6.48\)

(b) \(P(-\frac{4}{\sigma}<Z<\frac{4}{\sigma})=P(-0.6176<Z<0.6176)\)

\(\Phi=0.7317\)

Prob = \(2\Phi-1=2(0.7317)-1\)

= 0.463

### Question

The score when two fair six-sided dice are thrown is the sum of the two numbers on the upper faces.

(a) Show that the probability that the score is 4 is \(\frac{1}{12}\) [1]

The two dice are thrown repeatedly until a score of 4 is obtained. The number of throws taken is

denoted by the random variable X.

(b) Find the mean of X. [1]

** (c) Find the probability that a score of 4 is first obtained on the 6th throw. [1]**

(d) Find P(X < 8). [2]

**Answer/Explanation**

Ans

1 (a) Prob of 4 (from 1,3, 3,1 or 2,2) \(=\frac{3}{36}=\frac{1}{12} \ AG\)

1 (b) \(Mean=\frac{1}{\frac{1}{12}}=12\)

1 (c) \(\left ( \frac{11}{12} \right )\times \frac{1}{12}=0.0539\ or \frac{161051}{2985984}\)

1 (d) \(1-\left ( \frac{11}{12} \right )^{7}\)

\(0.456\ or \ \frac{16344637}{35831808}\ )

### Question

In a music competition, there are 8 pianists, 4 guitarists and 6 violinists. 7 of these musicians will be

selected to go through to the final.

How many different selections of 7 finalists can be made if there must be at least 2 pianists, at least

1 guitarist and more violinists than guitarists? [4]

**Answer/Explanation**

Ans

4 Scenarios:

2P 3V 2G ^{8}C_{2} × ^{4}C_{2}× ^{6}C_{3} = 28 × 6 × 20 = 3360

2P 4V 1G ^{8}C_{2} × ^{4}C_{1} × ^{6}C_{4} = 28 × 4 × 15 = 1680

3P 3V 1G ^{8}C_{3} × ^{4}C_{1}× ^{6}C_{3} = 56 × 4 × 20 = 4480

4P 2V 1G ^{8}C_{4} × ^{4}C_{1} × ^{6}C_{2} = 70 × 4 × 15 = 4200

(M1 for ^{8}C_{r} × ^{4}C_{r} × ^{6}C_{r} with Σr = 7

Two unsimplified products correct

Summing the number of ways for 3 or 4 correct scenarios

Total: 13 720

*Question*

Hannah chooses 5 singers from 15 applicants to appear in a concert. She lists the 5 singers in the order in which they will perform.

** (i)** How many different lists can Hannah make? [2]

Of the 15 applicants, 10 are female and 5 are male.

** (ii)** Find the number of lists in which the first performer is male, the second is female, the third is male, the fourth is female and the fifth is male. [2]

Hannah’s friend Ami would like the group of 5 performers to include more males than females. The order in which they perform is no longer relevant.

** (iii)** Find the number of different selections of 5 performers with more males than females. [3]

** (iv)** Two of the applicants are Mr and Mrs Blake. Find the number of different selections that include Mr and Mrs Blake and also fulfil Ami’s requirement. [3]

**Answer/Explanation**

Ans: