Probability Question #4

Probability Question #4

JC Mathematics

A gambler bets on one of the integers from 1 to 6. Three fair dice are then rolled. If the gambler’s number appears k times (k = 1, 2, 3), he wins $ k. If his number fails to appear, he loses $1. Calculate the gambler’s expected winnings

2016 A-level H2 Mathematics (9740) Paper 2 Suggested Solutions

2016 A-level H2 Mathematics (9740) Paper 2 Suggested Solutions

JC Mathematics

All solutions here are SUGGESTED. KS will hold no liability for any errors. Comments are entirely personal opinions.

Numerical Answers (click the questions for workings/explanation)

Question 1: 0.0251 \text{~m/min}
Question 2: \frac{x^2}{2} \text{sin}nx + \frac{2x}{n^2} \text{cos}nx - \frac{2}{n^3} \text{sin}nx + C;~ a = 2 \text{~or~} 6;~ \pi (\frac{2}{5} - \text{ln} \frac{3}{\sqrt{5}})
Question 3: a + \frac{3}{4} - \text{sin}a - \text{cos}a + \frac{1}{4} \text{cos}2a;~ \frac{1}{4}(\pi + 1)^2
Question 4: z = 2.63 + 1.93i, ~ 3.37 + 0.0715i ;~8^{\frac{1}{6}}e^{i(-\frac{\pi}{12})}, ~8^{\frac{1}{6}}e^{i(-\frac{3 \pi}{4})}, ~8^{\frac{1}{6}}e^{i(\frac{7\pi}{12})};~ n=7
Question 5: \frac{11}{42};~ \frac{4}{11};~ \frac{4}{1029}
Question 6: 60;~ 10;~ \{ \bar{x} \in \mathbb{R}, 0 \textless \bar{x} \le 34.8 \};~ \{ \alpha \in \mathbb{R}, 0 \textless \alpha \le 8.68 \}
Question 7: 24;~ 576;~ \frac{1}{12};~ \frac{5}{12}
Question 8: r = -0.980;~ c = -17.5;~ d = 91.8;~ y = 85.9
Question 9: a = 7.41;~ p = 0.599 \text{~or~} 0.166;~ 0.792
Question 10: 0.442;~ 0.151 ;~ 0.800;~ \lambda = 1.85

 

Relevant materials

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KS Comments

2015 A-level H2 Mathematics (9740) Paper 2 Question 11 Suggested Solutions

JC Mathematics

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Number of ways = \frac{8!}{2!2!} = 10080 ways

(ii)
Number of ways = 10080 -1 = 10079 ways

(iii)
Number of ways = 6! = 720 ways

(iv)
Case 1: 2 A’s together and B’s separated
5! \times ^6 C_2 = 1800
Case 2: 2 B’s together and A’s separated
5! \times ^6 C_2 = 1800
Case 3: 2 A’s together and 2 B’s together
720

Number of ways =10080 – 1800 – 1800 – 720 = 5760 ways

2015 A-level H2 Mathematics (9740) Paper 2 Question 8 Suggested Solutions

JC Mathematics

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

Let X denote the mass of pineapple tarts and \mu denote the population mean mass of pineapple tarts.

H_0: \mu = 0.9

H_1: \mu ~ \textless ~ 0.9

Unbiased estimate of population mean = 0.8825

s^2 = 0.0747854073^2 \approx 0.00559 (3 SF)

Under H_0, \bar{X} \sim \mathrm{N} (\mu, \frac{s^2}{n})

Test Statistic, T = \frac{\bar{X}- \mu}{\frac{s}{\sqrt{n}}} \sim t(7) at 10% level of significance.

Using GC, p-value = 0.264619 > 0.1

\Rightarrow, do not reject H_0

There is insufficient evidence at 10% significance level to reject the stall owner’s claim.

2015 A-level H2 Mathematics (9740) Paper 2 Question 6 Suggested Solutions

JC Mathematics

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Let X denote the number of red sweets in a small packet of 10 sweets.

X \sim \mathrm{Bin} (10, 0.25)

\mathrm{P}(X \ge 4) = 1 - \mathrm{P}(X \le 3) \approx 0.224 (3 SF)

(ii)
Let Y denote the number of red sweets in a large packet of 100 sweets.

Y \sim \mathrm{Bin} (100, 0.25)

Since n is large, np = 25 > 5, n(1-p) = 75 > 5

Y \sim \mathrm{N}(25, 18.75) approximately

\mathrm{P}( Y \ge 30) = \mathrm{P}(X > 29.5) by continuity correction
\approx 0.149 (3 SF)

(iii)
Let W denote the number of packets out of 15 packets that contain at least 30 red sweets.

Y \sim \mathrm{Bin}(15, 0.1493487984)

\mathrm{P}( W \le 3) \approx 0.825 (3 SF)