5 Marks Questions

Question 15 Marks

A factory has three machines A, B and C producing 1500, 2500 and 3000 bulbs per day, respectively. Machine A produces 1.5% defective bulbs, machine B produces 2% defective bulbs and machine C produces 2.5% defective bulbs. At the end of the day, a bulb is drawn at random and is found to be defective. What is the probability that the defective bulb has been produced by machine B?

Answer

Given that, total daily production of bulbs = 1500 + 2500 + 3000 = 7000 bulbs.
Let A, B and C be the event if drawing a bulb produced by machines A, B and C respectively.
$\therefore \quad P(A)=\frac{1500}{7000}=\frac{3}{14}$
$P(B)=\frac{2500}{7000}=\frac{5}{14}$
and $P(C)=\frac{3000}{7000}=\frac{3}{7}$
Let F be the event of producing a defective bulb
$P\left(\frac{F}{A}\right)=1.5 \%=\frac{1.5}{100}=\frac{3}{200}$
$P\left(\frac{F}{B}\right)=2 \%=\frac{2}{100}=\frac{1}{50}$
$P\left(\frac{F}{C}\right)=2.5 \%=\frac{2.5}{100}=\frac{1}{40}$
By Bayes' theorem,
$P\left(\frac{B}{F}\right)=\frac{P(B) \cdot P\left(\frac{F}{B}\right)}{P(A) \cdot P\left(\frac{F}{A}\right)+P(B) \cdot P\left(\frac{F}{B}\right)}$$+P(C) \cdot P(F / C)$
$=\frac{\frac{5}{14} \times \frac{1}{50}}{\frac{5}{14} \times \frac{1}{50}+\frac{3}{14} \times \frac{3}{200}+\frac{3}{7} \times \frac{1}{40}}$
$=\frac{\frac{1}{140}}{\frac{1}{140}+\frac{9}{2800}+\frac{3}{280}}=\frac{20}{59}$

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Question 25 Marks

An insurance company insured 1500 scooter drivers, 2500 car drivers and 4500 truck drivers. The probability of a scooter, a car and a truck meeting with an accident is 0.01, 0.02 and 0.04 respectively. If one of the insured persons meets with an accident, find the probability that he is a scooter driver.

Answer

Let $E_1, E_2$ and $E_3$ be event of insured driver of scooter, car and truck, respectively.
Then, $P\left(E_1\right)=\frac{1500}{8500}=\frac{15}{85}=\frac{3}{17}$
$P\left(E_2\right)=\frac{2500}{8500}=\frac{5}{17}$
and $P\left(E_3\right)=\frac{4500}{8500}=\frac{9}{17}$
Now, probability that the drivers meets with an accident has scooter driver.
$P\left(\frac{A}{E_1}\right)=0.01$
Probability that the drivers meets with an accidents has a car driver $P\left(\frac{A}{E_2}\right)=0.02$
Probability that the drivers meets with an accident has a truck drivers, $P\left(\frac{A}{E_3}\right)=0.04$
The probability of scooter driver meets with an accident,
$\begin{array}{c}P\left(\frac{E_1}{A}\right)=\frac{P\left(E_1\right) \cdot P\left(\frac{A}{E_1}\right)}{P\left(E_1\right) \cdot P\left(\frac{A}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{A}{E_2}\right)} \\ +P\left(E_3\right) \cdot P\left(\frac{A}{E_3}\right)\end{array}$
(by using Bayes' thorem)
$=\frac{\frac{3}{17} \times 0.01}{\frac{3}{17} \times 0.01+\frac{5}{17} \times 0.02+\frac{9}{17} \times 0.04}$
$=\frac{\frac{3}{17} \times 0.01}{\frac{1}{17}(3 \times 0.01+5 \times 0.02+9 \times 0.04)}$
$=\frac{3 \times 0.01}{0.03+0.10+0.36}$
$=\frac{0.03}{0.10+0.39}=\frac{0.03}{0.49}=\frac{3}{49}$

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Question 35 Marks

A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn at random and are found to be hearts. Find the probability of the missing card to be a heart.

Answer

Let $C_1, C_2, C_3$ and $C_4$ be the events that the lost card is of heart, spade, diamond or club, respectively.
Clearly, $P\left(C_1\right)=P\left(C_2\right)=P\left(C_3\right)=P\left(C_4\right)=\frac{13}{52}=\frac{1}{4}$
Let $S$ be the event of drawing two cards of heart from the remaining 51 cards. Here, we wish to find $P\left(\frac{C_1}{S}\right)$
Now, $P\left(\frac{S}{C_1}\right)$ is the probability of drawing two heart cards from 51 cards given that one heart card is lost
$=\frac{{ }^{12} C_2}{{ }^{51} C_2}=\frac{12 \times 11}{1 \times 2} \times \frac{1 \times 2}{51 \times 50}$
$=\frac{22}{425}$
Similarly,
$P\left(\frac{S}{C_2}\right)=P\left(\frac{S}{C_3}\right)=P\left(\frac{S}{C_4}\right)$
$=\frac{{ }^{13} C_2}{{ }^{51} C_2}=\frac{13 \times 12}{1 \times 2} \times \frac{1 \times 2}{51 \times 50}$
$=\frac{26}{425}$
By Bayes' Theorem,
$\begin{aligned} P\left(\frac{C_1}{S}\right)= & \frac{P\left(C_1\right) \cdot P\left(\frac{S}{C_1}\right)}{P\left(C_1\right) P\left(\frac{S}{C_1}\right)+P\left(C_2\right) \cdot P\left(\frac{S}{C_2}\right)} \\ & +P\left(C_3\right) \cdot P\left(\frac{S}{C_3}\right)+P\left(C_4\right) \cdot P\left(\frac{S}{C_4}\right)\end{aligned}$
$=\frac{\frac{1}{4} \times \frac{22}{425}}{\frac{1}{4} \times \frac{22}{425}+\frac{1}{4} \times \frac{26}{425}+\frac{1}{4} \times \frac{26}{425}+\frac{1}{4} \times \frac{26}{425}}$
$=\frac{22}{22+26+26+26}=\frac{11}{50}$

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Question 45 Marks

Bag A contains 2 white, 1 black and 3 red balls. Bag B contains 3 white, 2 black and 4 red balls and Bag C contains 4 white, 3 black and 2 red balls.
One bag is chosen at random and 2 balls are drawn at random from that bag. If the randomly drawn balls happen to be red and black, what is the probability that both come from Bag B?

Answer

Bag A
2	W
1	B
3	R

Bag B
3	W
2	B
4	R

Bag C
4	W
3	B
2	R

Let $E_1, E_2$ and $E_3$ be the events of choosing bag A, bag B and bag C, respectively.
$\therefore \quad P\left(E_1\right)=P\left(E_2\right)=P\left(E_3\right)=\frac{1}{3}$
Let E be the event of drawing 1 red and 1 black ball from the bag, then
$P\left(\frac{E}{E_1}\right)=$ Probability
(1 red and 1 black balls are drawn from Bag A
$=\frac{{ }^3 C_1 \times{ }^1 C_1}{{ }^6 C_2}=\frac{3}{15}=\frac{1}{5}$
Similarly
$P\left(\frac{E}{E_2}\right)=$ Probability
(1 red and 1 black balls are drawn from Bag B)
$=\frac{{ }^4 C_1 \times{ }^2 C_1}{{ }^9 C_2}=\frac{4 \times 2}{36}=\frac{2}{9}$
and $P\left(\frac{E}{E_3}\right)=\text { Probability }$
(1 red and 1 black balls are drawn from Bag C)
$=\frac{{ }^2 C_1 \times{ }^3 C_1}{{ }^9 C_2}=\frac{2 \times 3}{36}=\frac{1}{6}$
Required Probability $=$ Probability [2 balls (1 red and 1 black) are drawn from bag $B]$ is given by$P\left(\frac{E_2}{E}\right)$
$=\frac{P\left(E_2\right) \cdot P\left(\frac{E}{E_2}\right)}{P\left(E_1\right) \cdot P\left(\frac{E}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{E}{E_2}\right)+P\left(E_3\right) \cdot P\left(\frac{E}{E_3}\right)}$
$=\frac{\frac{1}{3} \times \frac{2}{9}}{\frac{1}{3} \times \frac{1}{5}+\frac{1}{3} \times \frac{2}{9}+\frac{1}{3} \times \frac{1}{6}}$
$=\frac{\frac{1}{3} \times \frac{2}{9}}{\frac{1}{3}\left[\frac{1}{5}+\frac{2}{9}+\frac{1}{6}\right]}=\frac{\frac{2}{9}}{\frac{18+20+15}{90}}=\frac{2 \times 90}{9 \times 53}$
$=\frac{20}{53}$

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Applied Maths 5 Marks Questions

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