SPM 2023 Add Maths Paper 1, Question 11Probability Distribution

Solved on video in English and Bahasa Melayu · 3 marks · Form 5, Probability Distribution · 5:35 video

The question

(a)

It is found that the probability that a product from a company can be sold is 19/20. If in a particular month the company produces 820 units of the product, find the mean and the standard deviation of the number of products expected to be sold that month.

[3 marks]

(b)(i)

There are 25 students in a class. A teacher wants to choose some students from the class to join a game. It is given that the number of different ways of choosing r students is equal to the number of different ways of choosing (r + 13) students. (b)(i) Find the value of r.

(b)(ii)

The teacher decides to choose 8 students to join the game, with equal numbers of boys and girls (4 boys and 4 girls). If these 8 students are to be arranged in a row, find the number of different ways such that the boys cannot be next to each other.

SPM 2023 Add Maths Paper 1, Question 11: the question as drawn in Eduly's video5:35

Part (a)

Step 1

Binomial distribution with p = 19/20 = 0.95, q = 1 − p = 1/20 = 0.05, n = 820.

Why this step

Each unit is either sold or not sold, with the same chance each time and a fixed number of units. That pattern tells us it's binomial.

2 more steps, each with the reason for it, in the full solution

Final answer

Mean = 779, standard deviation = 6.241

Where students lose marks

Watch out: npq = 38.95 is the variance. Take the square root to get the standard deviation.

Part (b)(i)

3 more steps, each with the reason for it, in the full solution

Final answer

r = 6

Where students lose marks

Watch out: don't expand factorials and get stuck. Use nC_a = nC_b when a + b = n to form the equation.

Part (b)(ii)

4 more steps, each with the reason for it, in the full solution

Final answer

2880 ways

Where students lose marks

Watch out: listing just a few boy-girl patterns misses many valid ones (giving 1152). Use the gap method to count them all.

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