Chapter 13: Probability

Ans: It is given in the question that at random one ball is picked and its colour is noted then two balls of same colour is added, hence we have two cases here 

Consider the first case

Let the first ball drawn be Red, its probability is $\frac{5}{10}$

Now two extra red balls are added, 

$\therefore$ P (drawing red ball at second attempt =$\frac{7}{12}$

Since these two events are independent therefore required probability for this case is $\frac{5}{10}\times \frac{7}{12}$

Consider the second case

Let black ball be drawn at first attempt, its probability is $\frac{5}{10}$

Now two extra black balls will be added 

 $\therefore$ P(drawing red ball at second attempt)=$\frac{5}{12}$

Since these two events are independent therefore required probability for this case is $\frac{5}{10}\times \frac{5}{12}$

Total probability is given by $\frac{5}{10}\times \frac{7}{12}+\frac{5}{10}\times \frac{5}{12}=\frac{5}{10}$

Thus the probability of drawing second ball red is $\frac{1}{2}$

Ans: Let E and F be the events of selecting first and second bag respectively

$\therefore P\left( E \right)=P\left( F \right)=\frac{1}{2}$

Let K be the event of getting a red ball 

$\therefore P\left( K\left| E \right. \right)=\frac{4}{8}$, where  

$\left( K\left| E \right. \right)$: drawing red ball from the first bag

Similarly, $P\left( K\left| F \right. \right)=\frac{2}{8}$, where  

$\left( K\left| F \right. \right)$: drawing red ball from the first bag

Now from the Bay’s theorem 

$P\left( E\left| K \right. \right)=\frac{P\left( E \right)P\left( K\left| E \right. \right)}{P\left( E \right)P\left( K\left| E \right. \right)+P\left( F \right)P\left( K\left| F \right. \right)}$

\[\Rightarrow P\left( E\left| K \right. \right)=\frac{\frac{1}{2}\times \frac{1}{2}}{\frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{4}}\]

Thus the probability that the red ball is drawn from the first bag is$\frac{2}{3}$ 

Ans: Let us have following notations

E: The event when a student is a hostler

F:  The event when a student is a day scholar

K: The event chosen student gets grade A

$\therefore P\left( E \right)=0.6$

$\therefore P\left( F \right)=0.4$

$\therefore P\left( K\left| E \right. \right)=0.3$

$\therefore P\left( K\left| F \right. \right)=0.2$

By Bay’s theorem 

$P\left( E\left| K \right. \right)=\frac{P\left( E \right)P\left( K\left| E \right. \right)}{P\left( E \right)P\left( K\left| E \right. \right)+P\left( F \right)P\left( K\left| F \right. \right)}$

\[\Rightarrow P\left( E\left| K \right. \right)=\frac{0.6\times 0.3}{0.6\times 0.3+0.4\times 0.2}\]

Thus the probability that the gets grade A provided he Is a hostler is $\frac{9}{13}$ 

Ans: Let us have following notations

E: The event when a student knows the answer

F:  The event when a student guesses the answer

K: The event that the answer is correct

$\therefore P\left( E \right)=\frac{3}{4}$

$\therefore P\left( F \right)=\frac{1}{4}$

$\therefore P\left( K\left| E \right. \right)=1$ (since if he knows the answer then he will answer it correctly, it’s a trivial case)

where  $\left( K\left| E \right. \right)$: the answer is correct provided he knows the answer 

$\therefore P\left( K\left| F \right. \right)=\frac{1}{4}$

$\left( K\left| F \right. \right)$: the answer is correct provided he guesses the answer 

By Bay’s theorem 

$P\left( E\left| K \right. \right)=\frac{P\left( E \right)P\left( K\left| E \right. \right)}{P\left( E \right)P\left( K\left| E \right. \right)+P\left( F \right)P\left( K\left| F \right. \right)}$

\[\Rightarrow P\left( E\left| K \right. \right)=\frac{\frac{3}{4}\times \frac{1}{1}}{\frac{3}{4}\times \frac{1}{1}+\frac{1}{4}\times \frac{1}{4}}\]

Thus the probability that he knows the answer provided he answered correctly is $\frac{12}{13}$ 

Ans: Let us have following notations

E: The event when the person has disease

F:  The event when the person does not have disease

K: The event that the test result is positive

$\therefore P\left( E \right)=0.001$

It is clear that E and F are complementary events 

$\therefore P\left( E \right)+P\left( F \right)=1$

Hence 

$P\left( F \right)=0.999$

$\therefore P\left( K\left| E \right. \right)=0.99$ 

$\left( K\left| E \right. \right)$: the result is correct provided he has the disease 

$\therefore P\left( K\left| F \right. \right)=0.005$

$\left( K\left| F \right. \right)$: the result is correct provided he doesn’t have the disease

By Bay’s theorem 

$P\left( E\left| K \right. \right)=\frac{P\left( E \right)P\left( K\left| E \right. \right)}{P\left( E \right)P\left( K\left| E \right. \right)+P\left( F \right)P\left( K\left| F \right. \right)}$

\[\Rightarrow P\left( E\left| K \right. \right)=\frac{0.001\times 0.99}{0.001\times 0.99+0.999\times 0.005}\]

Thus the probability that the person had the disease provided his test result is positive is $\frac{22}{133}$

Ans: Let us have following notations

E: The event when chosen coin is two-headed 

F:  The event when chosen coin is biased 

G: The event when chosen coin is unbiased

K: The event that the coin shows heads

$\therefore P\left( E \right)=\frac{1}{3}$

$P\left( F \right)=\frac{1}{3}$

$P\left( G \right)=\frac{1}{3}$

$\therefore P\left( K\left| E \right. \right)=1$  (since it is a trivial case)

$\left( K\left| E \right. \right)$: the coins shows heads provided it is two headed 

$\therefore P\left( K\left| F \right. \right)=\frac{3}{4}$

$\left( K\left| F \right. \right)$: the coins shows heads provided it is biased

$\therefore P\left( K\left| G \right. \right)=\frac{1}{2}$

$\left( K\left| G \right. \right)$: the coins shows heads provided it is unbiased

By Baye’s theorem 

$P\left( E\left| K \right. \right)=\frac{P\left( E \right)P\left( K\left| E \right. \right)}{P\left( E \right)P\left( K\left| E \right. \right)+P\left( F \right)P\left( K\left| F \right. \right)+P\left( G \right)P\left( K\left| G \right. \right)}$

\[P\left( E\left| K \right. \right)=\frac{\frac{1}{3}\times 1}{\frac{1}{3}\times 1+\frac{1}{3}\times \frac{3}{4}+\frac{1}{3}\times \frac{1}{2}}\]

Thus the probability that the coin shows heads provided it is two headed coin is $\frac{4}{9}$

Ans: Let us have following notations

E: The event when the driver drives scooter 

F:  The event when the driver drives car

G: The event when the driver drives truck

K: The event that the driver meets accident

$\therefore P\left( E \right)=\frac{2000}{12000}$

$P\left( F \right)=\frac{4000}{12000}$

$P\left( G \right)=\frac{6000}{12000}$

$\therefore P\left( K\left| E \right. \right)=\frac{1}{100}$  

$\left( K\left| E \right. \right)$: the driver meets accident provided he drives scooter

$\therefore P\left( K\left| F \right. \right)=\frac{3}{100}$

$\left( K\left| F \right. \right)$: the driver meets accident provided he drives car

$\therefore P\left( K\left| G \right. \right)=\frac{15}{100}$

$\left( K\left| G \right. \right)$: the driver meets accident provided he drives truck

By Baye’s theorem 

$P\left( E\left| K \right. \right)=\frac{P\left( E \right)P\left( K\left| E \right. \right)}{P\left( E \right)P\left( K\left| E \right. \right)+P\left( F \right)P\left( K\left| F \right. \right)+P\left( G \right)P\left( K\left| G \right. \right)}$

\[P\left( E\left| K \right. \right)=\frac{\frac{1}{6}\times \frac{1}{100}}{\frac{1}{6}\times \frac{1}{100}+\frac{1}{3}\times \frac{3}{100}+\frac{1}{2}\times \frac{15}{100}}\]

Thus the probability that the accidented person is a scooter driver is $\frac{1}{52}$

Ans: Let us have following notations

E: The event when the production is from machine A

F:  The event when the production is from machine B

K: The event that the item is defective

$\therefore P\left( E \right)=0.6$ 

$\therefore P\left( F \right)=0.4$

$\therefore P\left( K\left| E \right. \right)=0.02$ 

$\left( K\left| E \right. \right)$: The event that the item is defective provided it from A

$\therefore P\left( K\left| F \right. \right)=0.01$

$\left( K\left| F \right. \right)$: The event that the item is defective provided it from B

By Baye’s theorem 

$P\left( F\left| K \right. \right)=\frac{P\left( F \right)P\left( K\left| F \right. \right)}{P\left( E \right)P\left( K\left| E \right. \right)+P\left( F \right)P\left( K\left| F \right. \right)}$

\[\Rightarrow P\left( F\left| K \right. \right)=\frac{0.4\times 0.01}{0.6\times 0.02+0.4\times 0.01}\]

Thus the probability that that the item is defective provided it from B is $\frac{1}{4}$

Ans: Let us have following notations

E: The event when the first group wins

F:  The event when the second group wins

K: The event that the new item is produced

$\therefore P\left( E \right)=0.6$ 

$\therefore P\left( F \right)=0.4$

$\therefore P\left( K\left| E \right. \right)=0.7$ 

$\left( K\left| E \right. \right)$: The event that the item is introduced by first group 

$\therefore P\left( K\left| F \right. \right)=0.3$

$\left( K\left| F \right. \right)$: The event that item is introduced by second group

By Baye’s theorem 

$P\left( F\left| K \right. \right)=\frac{P\left( F \right)P\left( K\left| F \right. \right)}{P\left( E \right)P\left( K\left| E \right. \right)+P\left( F \right)P\left( K\left| F \right. \right)}$

\[\Rightarrow P\left( F\left| K \right. \right)=\frac{0.4\times 0.3}{0.6\times 0.7+0.4\times 0.3}\]

Thus the probability that that the item is produced from second group is $\frac{2}{9}$

Ans: Let us have following notations

E: The event when $5\text{ or 6}$ come

F:  The event when $\text{1,2,3 or 4}$ come

K: The event that the coin shows exactly one head

$\therefore P\left( E \right)=\frac{2}{6}$ 

$\therefore P\left( F \right)=\frac{4}{6}$

$\therefore P\left( K\left| E \right. \right)=\frac{3}{8}$ 

$\left( K\left| E \right. \right)$: The event that the coin shows exactly one head provided $5\text{ or 6}$come

$\therefore P\left( K\left| F \right. \right)=\frac{1}{2}$

$\left( K\left| F \right. \right)$: The event that the single throw of coin shows exactly one head provided $\text{1,2,3 or 4}$come

By Baye’s theorem 

$P\left( F\left| K \right. \right)=\frac{P\left( F \right)P\left( K\left| F \right. \right)}{P\left( E \right)P\left( K\left| E \right. \right)+P\left( F \right)P\left( K\left| F \right. \right)}$

\[\Rightarrow P\left( F\left| K \right. \right)=\frac{\frac{4}{6}\times \frac{1}{2}}{\frac{2}{6}\times \frac{3}{8}+\frac{4}{6}\times \frac{1}{2}}\]

Thus the probability that that the item is produced from second group is $\frac{8}{11}$