Question 1
Topic – ALV: 5.3
Two fair coins are thrown at the same time repeatedly until a pair of heads is obtained. The number of throws taken is denoted by the random variable X.
(a) State the value of E(X).
(b) Find the probability that exactly 5 throws are required to obtain a pair of heads.
(c) Find the probability that fewer than 7 throws are required to obtain a pair of heads.
▶️Answer/Explanation
Solution:-
1(a) $\left[P(HH) = \frac{1}{4}\right]$ $\left[E(X) = 4\right]$
1(b) $\left[P(X = 5) = \left(\frac{3}{4}\right)^4 \left(\frac{1}{4}\right)\right] = 0.0791$
1(c) $\left[P(X < 7) = 1 – \left(\frac{3}{4}\right)^6\right]$
or $\frac{1}{4} + \frac{1}{4} \times \frac{3}{4} + \frac{1}{4} \times \left(\frac{3}{4}\right)^2 + … + \frac{1}{4} \times \left(\frac{3}{4}\right)^5 – \frac{1}{4}$
$= \frac{3367}{4096}, 0.822$
Question 2
Topic – ALV: 5.3
Anil is a candidate in an election. He received 40% of the votes. A random sample of 120 voters is chosen.
Use an approximation to find the probability that, of the 120 voters, between 36 and 54 inclusive voted for Anil.
▶️Answer/Explanation
Solution:-
$Mean = 120 \times 0.4 = 48$
$Var = 120 \times 0.4 \times 0.6 = 28.8$
$P(36 \leq X \leq 54) = P(\frac{35.5 – 48}{\sqrt{28.8}} < Z < \frac{54.5 – 48}{\sqrt{28.8}})$
$[=P(-2.3292 < Z < 1.211) =] 0.8871 + 0.9900 – 1$
= 0.877
Question 3
Topic – ALV: 5.4
The random variable X takes the values 1, 2, 3, 4. It is given that P(X = x) = kx(x + a), where k and a are constants.
(a) Given that P(X = 4) = 3P(X = 2), find the value of a and the value of k.
(b) Draw up the probability distribution table for X, giving the probabilities as numerical fractions.
(c) Given that E(X) = 3.2, find Var(X).
▶️Answer/Explanation
Solution:-
(a) $[P(X=4)=3P(X=2)]$
$4k(4+a)=3 \times 2k(2+a)$
$16k+4ak=12k+6ak$
$a=2$
$3k+8k+15k+24k=1$
$k=\frac{1}{50}$
(b)
(c) $Var(X) = \frac{3}{50} \times 1^2 + \frac{8}{50} \times 2^2 + \frac{15}{50} \times 3^2 + \frac{24}{50} \times 4^2 – 3.2^2$
$[= 11.08 – 3.2^2 =] 0.84 \left[0\right], \frac{21}{25}$
Question 4
Topic – ALV: 5.1
The times taken, in minutes, to complete a cycle race by 19 cyclists from each of two clubs, the Cheetahs and the Panthers, are represented in the following back-to-back stem-and-leaf diagram.
(a) Find the median and the interquartile range of the times of the Cheetahs.
The median and interquartile range for the Panthers are 103 minutes and 14 minutes respectively.
(b) Make two comparisons between the times taken by the Cheetahs and the times taken by the Panthers.
Another cyclist, Kenny, from the Cheetahs also took part in the race. The mean time taken by the 20 cyclists from the Cheetahs was 99 minutes.
(c) Find the time taken by Kenny to complete the race.
▶️Answer/Explanation
Solution:-
(a) $Median = 99 [minutes]$
$[IQR=]106-83$
23 [minutes]
(b) The times for the Cheetahs are faster than the times for the Panthers
The times for the Cheetahs are more spread than the times for the Panthers
(c) $[Total time including Kenny = 99 \times 20 =]1980$
$[Kenny’s time =] 1980-1862$
= 118 [minutes]
Alternative Method for Question $4(c)$
1862 + their Kenny’s time = 99
20
$[Kenny’s time = 99 \times 20-1862]$
= 118 [minutes]
Question 5
Topic – ALV: 5.3
Jasmine throws two ordinary fair 6-sided dice at the same time and notes the numbers on the uppermost faces. The events A and B are defined as follows.
A: The sum of the two numbers is less than 6.
B: The difference between the two numbers is at most 2.
(a) Determine whether or not the events A and B are independent.
(b) Find P(B | A’).
▶️Answer/Explanation
Solution:-
(a) $P(A) = \frac{10}{36}$
$P(B) = \frac{24}{36}$
$P(A \cap B) = \frac{8}{36}$
$\frac{10}{36} \times \frac{24}{36}$
$= \frac{5}{27}, 0.185 \left[ \neq \frac{8}{36} \right]$
Events are not independent
(b) $\left[P(B|A’) = \frac{P(B \cap A’)}{P(A’)}\right] = \frac{\frac{16}{36}}{\left(1 – \frac{10}{36}\right)}$
= $\frac{8}{13}$
Alternative Method for Question 5(b): Direct from outcome tables
$\left[P(B|A’) = \frac{Number \text{ } of \text{ } outcomes \text{ } (B \cap A’)}{Number \text{ } of \text{ } outcomes \text{ } (A’)}\right] = \frac{16}{26}$
Question 6
Topic – ALV: 5.5
The mass of grapes sold per day by a large shop can be modelled by a normal distribution with mean 28kg. On 10% of days less than 16kg of grapes are sold.
(a) Find the standard deviation of the mass of grapes sold per day.
The mass of grapes sold on any day is independent of the mass sold on any other day.
(b) 12 days are chosen at random.
Find the probability that less than 16kg of grapes are sold on more than 2 of these 12 days.
(c) In a random sample of 365 days, on how many days would you expect the mass of grapes sold to be within 1.3 standard deviations of the mean?
▶️Answer/Explanation
Solution:-
(a) $[P(X < 16) = P(Z < \frac{16 – 28}{\sigma}) = 0.1]$
$\frac{16-28}{\sigma} = -1.282$
$\sigma = 9.36$
(b) $[1 – P(0, 1, 2) =] 1 – (^{12}C_{0}(0.1)^{0}(0.9)^{12} + ^{12}C_{1}(0.1)^{1}(0.9)^{11} + ^{12}C_{2}(0.1)^{2}(0.9)^{10})$
$[1 – (0.2824 + 0.3766 + 0.2301)]$
0.111
Alternative Method for Question 6(b)
$P(3, 4, 5, 6, 7, 8, 9, 10, 11, 12) = ^{12}C_{3}(0.1)^{3}(0.9)^{9} + ^{12}C_{4}(0.1)^{4}(0.9)^{8} + … + ^{12}C_{11}(0.1)^{11}(0.9)^{1} + ^{12}C_{12}(0.1)^{12}(0.9)^{0}$
$[0.08523 + 0.02131 + … + 1.08 \times 10^{-10} + 1 \times 10^{-12}]$
0.111
(c) $[P(-1.3 < Z < 1.3)]$
$= 2 \Phi(1.3) – 1]$
$= 2 \times 0.9032 – 1$
$= 0.806, \frac{504}{625}$
$[In 365 days 0.8064 \times 365]$
= 294 or 295
Question 7
Topic – ALV: 5.2
(a) Find the number of different arrangements of the 10 letters in the word CASABLANCA in which the two Cs are not together.
(b) Find the number of different arrangements of the 10 letters in the word CASABLANCA which have an A at the beginning, an A at the end and exactly 3 letters between the 2 Cs.
Five letters are selected from the 10 letters in the word CASABLANCA.
(c) Find the number of different selections in which the five letters include at least two As and at most one C.
▶️Answer/Explanation
Solution:-
(a) $\quad \text{Method 1: Total number of arrangements – number of arrangements with Cs together}$
$\frac{10!}{2!4!} – \frac{9!}{4!} [75600-15120]$
= 60480
Method 2: Arrangements ^^^^^^^
$\frac{8!}{4!} \times \frac{9 \times 8}{2}$
= 60480
(b) $\text{A C C C C C A A A}$
$\frac{6!}{2!} \times 4$
1440
Alternative Method for Question 7(b)
$\frac{4 \times ^6P_{3} \times 3!}{2!}$
1440
(c)$ \text{Scenarios:}$
AA — ${}^{5}C_{3}=10$
AAA — ${}^{5}C_{2}=10$
AAAA — ${}^{5}C_{1}=5$
25
Alternative Method 2 for Question 7(c)
Scenarios:
AAC — ${}^{4}C_{2}=6$
AA — ${}^{4}C_{3}=4$
AAA — ${}^{4}C_{2}=6$
AAAA — ${}^{4}C_{1}=4$
1
4
25