The availability of step-by-step ML Aggarwal Maths for Class 12 Solutions Chapter 10 Probability Ex 10.4 can make challenging problems more manageable.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 10 Probability Ex 10.4

Question 1.

A bag contains 3 white and 6 black balls while another bag contains 6 white and 3 black balls. A bag is selected at random and a ball is drawn. Find the probability that the ball drawn is of white colour.

Answer:

Let E_{1}, E_{2} and A be the events defined as follows :

E_{1} : event that first bag is chosen

E_{2} : event that second bag is selected A : event that ball drawn be white

Here E_{1} and E_{2} are mutually exclusive and exhausitive events.

P(E_{1}) = P(E_{2}) = \(\frac{1}{2}\)

P (A/E_{1}) = prob. of drawing a white ball from bag first = \(\frac{3}{9}=\frac{1}{3}\)

P (A/E_{2}) = prob. of draiving’a white ball from bag second = \(\frac{6}{9}=\frac{2}{3}\)

Then by using law of total probability ; we have

P (A) = P (E_{1}) P (A/E_{1}) + P (E_{2}) P (A/E_{2})

= \(\frac{1}{2} \times \frac{1}{3}+\frac{1}{2} \times \frac{2}{3}=\frac{1}{6}+\frac{1}{3}=\frac{3}{6}=\frac{1}{2}\)

Question 2.

A purse contains 4 silver and 5 copper coins. A second purse contains 3 silver and 7 copper coins. If a coin is taken out at random from one of the purses, what is the probability that it is a copper coin ? (ISC 2011)

Answer:

Let E_{1}, E_{2} and A be the events defined as follows :

E_{1} : event that first purse is chosen

E_{2} : event that second purse is chosen A : event that coin drawn as copper coin.

Since E_{1} and E_{2} are mutually exclusive and exhaustive events

P(E_{1}) = P(E_{2}) = \(\frac{1}{2}\)

P (A/E_{1}) = prob. of drawing a copper coin from purse-I = \(\frac{5}{9}\)

P (A/E_{2}) = prob. ot drawing a copper coin from purse-II = \(\frac{7}{10}\)

Then by using law of total probability ; we have

P (A) = P (E_{1}) P (A/E_{1}) + P (E_{2}) P (A/E_{2})

= \(\frac{1}{2} \times \frac{5}{9}+\frac{1}{2} \times \frac{7}{10}=\frac{50+62}{180}=\frac{113}{180}\)

Question 3.

Suppose that 5 men out of 100 and 25 women out of 1000 are good orators. Assumiag that there are equal number of men and women, find the probability of choosing a good orator.

Answer:

Let E : event of getting a good orator

E_{1} : choosing a man

E_{2} : choosing a woman

Here P E/E_{1}) = \(\frac{5}{100}\); P (E/E_{2}) = \(\frac{25}{1000}\)

∴ P(E_{1}) = P(E_{2}) = \(\frac{1}{2}\)

Thus required prob.P(E) = P(E_{1})P(E/E_{1}) + P(E_{2}) + P(E/E_{2})

= \(\frac{1}{2} \times \frac{5}{100}+\frac{1}{2} \times \frac{25}{1000}\)

= \(\frac{75}{2000}=\frac{3}{80}\)

Question 4.

A person has undertaken a construction job. The probabilities are 0.65 that there will be strike, 0.80 that the construction job will be completed on time if there is no strike, and 0.32 that the construction job will be completed on time if there is a strike. Determine the probability that the construction job will be completed on time. How construction is affected in case of riots ? What values are being promoted ? (Value Based)

Answer:

Let S : event that there will be a strike

C : event that construction job will be completed in time given

P (S) = prob. that there will be a strike = 0.65

P(C/S) = prob. of construction job will be completed on time if there is a strike = 0.32

P(S)= 1 – P (S) = 1 – 0.65 = 0.35

P (C/ S)= prob. that construction job will be completed on time if there is no strike = 0.80

Then required probability = P(C)

= P(S)P(C/S)+ P(S) P(C/S)

= 0.65 × 0.32 + 0.35 × 0.80

= 0.488

Since Riots creates unnecessary tension between peoples of different religions. During strikes, there will be waste of man hours and there will be no productive activities.

Question 5.

There are two bags. One bag contains green and three red balls. The second bag contains five green and four red ? balls. One ball is transferred from the first bag to the second bag. Then one ball is drawn from the second bag. Find the probability that it is a red ball. (ISC 2010)

Answer:

Given, Bag I contains six green and three red balls

Bag II contains five green and four red balls

Let us define the events E_{1}, E_{2} and A as follows:

E_{1} : green ball is transferred from the first bag to second

E_{2} : red ball is transferred from first bag to second bag.

A : ball drawn from second bag be red ball.

Clearly E_{1} and E_{2} are mutually exclusive and exhaustive.

P(E_{1}) = \(\frac{6}{9}=\frac{2}{3}\)

P(E_{2}) = \(\frac{3}{9}=\frac{1}{3}\)

P(A/E_{1}) = \(\frac{4}{6+4}=\frac{4}{10}=\frac{2}{5}\)

[When a green ball is transferred from first bag to second then second bag contains 6 green and four red bals]

P (A/E_{2}) = \(\frac{5}{5+5}=\frac{5}{10}=\frac{1}{2}\)

[When a red ball transferred from bag I to bag II then second bag contains 5 green and 5 red balls]

P (A) = P (E_{1}) P (A/E_{1}) + P (E_{2}) P (A/E_{2})

= \(\frac{2}{3} \times \frac{2}{5}+\frac{1}{3} \times \frac{1}{2}=\frac{4}{15}+\frac{1}{6}\)

= \(\frac{8+5}{30}=\frac{13}{30}\)

Question 6.

An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is . noted and is then returned to the urn. C Moreover, 2 additional balls of the same colour are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red ? (NCERT)

Answer:

Given urn contains 5 red and 5 black balls.

Let E_{1}, E_{2} and A be the events defined as follows :

E_{1} : red ball drawn from urn in the first draw

E_{2} : black ball drawn from urn in the first draw

A : Red ball is drawn from the urn in second draw.

Since E_{1} and E_{2} are mutually exclusive and exhaustive events

P(E_{1}) = \(\frac{5}{5+5}=\frac{5}{10}=\frac{1}{2}\)

P(E_{2}) = \(\frac{5}{10}=\frac{1}{2}\)

P (A/E_{1}) = prob. of drawing a red ball from urn in second draw after putting 2 red balls into the urn

P (A/E_{2}) = prob. of draw a red ball from urn in second draw af^er putting 2 black balls into urn

Then by law of total probability ; we have

P (A) = P (E_{1}) P (A/E_{1}) + P (E_{2}) P (A/E_{2})

= \(\frac{1}{2} \times \frac{7}{12}+\frac{1}{2} \times \frac{5}{12}=\frac{12}{24}=\frac{1}{2}\)

Question 7.

There are two bags, one of which contains 3 black and 4 white balls while the other contains 4 black and 3 white balls. A die is thrown. If it shows up 1 or 3, a ball is taken from the first bag ; but if it shows up any other number, a ball is chosen from the second bag. Find the probability of choosing a black ball. (NCERT Exemplar)

Answer:

Given, bag I contains 3 black and 4 white balls and bag II contains 4 black and 3 white balls

Let E_{1}, E_{2} and A be the events defined as follows:

E_{1} : bag I is picked up

E_{2} : bag II is chosen

A : a black ball is drawn from selected bag

Since E_{1} and E_{2} are mutually exclusive and exhaustive events

Then P (E_{1}) = \(\frac{2}{6}=\frac{1}{3}\);

P (E_{2}) = \(\frac{4}{6}=\frac{2}{3}\)

P (A/E_{1}) = prob. of drawing a black ball when bag I is chosen

= \(\frac{3}{3+4}=\frac{3}{7}\)

A (A/E_{2}) = prob. of drawing a black ball when bag II is picked up

= \(\frac{1}{3} \times \frac{3}{7}+\frac{2}{3} \times \frac{4}{7}=\frac{11}{21}\)

Then by law of total probability ; we have

P (A) = P (E_{1}) P (A/E_{1}) + P (E_{2}) P (A/E_{2})

= \(\frac{1}{3} \times \frac{3}{7}+\frac{2}{3} \times \frac{4}{7}=\frac{11}{21}\)

Question 8.

Urn I has 2 white and 3 black balls, urn II has 4 white and 1 black ball and urn III has 3 white and 2 black balls. An urn & is selected at random and a ball is drawn at random. What is the probability of drawing a white ball ?

Answer:

Given urn I contains 2 white and 3 black balls

urn II contains 4 white and 1 black balls

urn III contains 3 white and 2 black balls

Let us define the events E_{1}, E_{2}, E_{3} and A are as follows :

E_{1} : urn I is chosen ; E_{2} : urn II is chosen

E_{3} : urn III is selected; A : white ball is drawn from selected urn

E_{1}, E_{2} and E_{3} are mutually exclusive and exhaustive events.

P (E_{1}) = P (E_{2}) = P (E_{3}) = \(\frac{1}{3}\)

P (A/E_{1}) = prob. of drawing white ball from urn I = \(\frac{2}{2+3}=\frac{2}{5}\)

P (A/E_{2}) = prob. of drawing white ball from urn II = \(\frac{4}{4+1}=\frac{4}{5}\)

P (A/E_{3}) = prob. of drawing white ball from urn III = \(\frac{3}{3+2}=\frac{3}{5}\)

Then by using law of total probability ; we have

P(A) = P (E_{1}) P (A/E_{1}) + P (E_{2}) P (A/E_{2}) + P (E_{3}) P (A/E_{3})

= \(\frac{1}{3} \times \frac{2}{5}+\frac{1}{3} \times \frac{4}{5}+\frac{1}{3} \times \frac{3}{5}\)

= \(\frac{2+4+3}{15}=\frac{9}{15}=\frac{3}{5}\)

Question 9.

In a factory, a product is manufactured by any of three machines A, B and C. They produce respectively 25%, 35% and 40% of the total products. A product is selected at random. Find the probability that it is defective. Assume that machines A, B and C produce respectively 5%, 4% and 2% defective items.

Answer:

Let E_{1}, E_{2}, E_{3} and A be the events defined as follows :

E_{1}: product is manufactured by machine A

E_{2} : product is manufactured by machine B

E_{3} : product is manufactured by machine C A : A defective item is selected

Thus E_{1}, E_{2}, E_{3} are mutually exclusive and exhaustive events

P(E_{1}) = \(\frac{25}{100}\); P(E_{2}) = \(\frac{35}{100}\); P(E_{3}) = \(\frac{40}{100}\)

P (A/E_{1}) = prob. of getting defective item from machine A = \(\frac{5}{100}\)

P (A/E_{2}) = probability of getting defective item from machine B = \(\frac{4}{100}\)

P (A/E_{3}) = probability of getting defective item from machine C = \(\frac{2}{100}\)

Then by law of total probability ; we have

P (A) = P (E_{1}) P (A/E_{1}) + P (E_{2}) P (A/E_{2}) + P (E_{3}) P (A/E_{3})

= \(\frac{25}{100} \times \frac{5}{100}+\frac{35}{100} \times \frac{4}{100}+\frac{40}{100} \times \frac{2}{100}\)

= \(\frac{125+140+80}{1000}=\frac{345}{10000}\)

= \(\frac{69}{2000}\)

Question 10.

Bag A contains 1 white, 2 blue and 3 red balls. Bag B contains 3 white, 3 blue and 2 red balls. Bag C contains 2 white, 3 blue and 4 red balls. One bag is selected at random, and \0 then two balls are drawn from the selected bag. Find the probability that the balls drawn are white and red. (ISC 2017)

Answer:

Given, bag A contains 1 white, 2 blue and 3 red balls bag B contains 3 white, 3 blue and 2 red balls and bag C contains 2 white, 3 blue and 4 red balls

Let E_{1} : event that bag A is chosen

E_{2} : event that bag B is chosen and

E_{3} : event that bag C is chosen

E_{1}, E_{2} and E_{3} are mutually exclusive and exhaustive events.

Then P (E_{1}) = P (E_{2}) = P (E_{3}) = \(\)

A : event that one white and one red ball drawn from selected bag

P (A/E_{1}) = probability of drawing one white and one red ball from bag A

= \(\frac{{ }^1 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1}{{ }^6 \mathrm{C}_2}=\frac{3 \times 2}{6 \times 5}=\frac{1}{5}\)

P (A/E_{2}) = prob. of drawing one white and one red ball from bag B

= \(\frac{{ }^3 \mathrm{C}_1 \times{ }^2 \mathrm{C}_1}{{ }^8 \mathrm{C}_2}=\frac{3 \times 2 \times 2}{8 \times 7}=\frac{3}{14}\)

P (A/E_{3}) = probability of drawing one white and one red ball from bag

= \(\frac{{ }^2 C_1 \times{ }^4 C_1}{{ }^9 C_2}=\frac{2 \times 4 \times 2}{9 \times 8}=\frac{2}{9}\)

Thus by using law of Total probability, P (A) = required probability

= P (E_{1}) P (A/E_{1}) + P (E_{2}) P (A/E_{2}) + P (E_{3}) P (A/E_{3})

= \(\frac{1}{3} \times \frac{1}{5}+\frac{1}{3} \times \frac{3}{14}+\frac{1}{3} \times \frac{2}{9}=\frac{1}{3}\left[\frac{1}{5}+\frac{3}{14}+\frac{2}{9}\right]\)

= \(\frac{1}{3}\left[\frac{126+135+140}{630}\right]=\frac{401}{1890}\)

Question 10 (Old).

An urn contains 2 white and 2 black balls. A ball is drawn at random. If it is white, it is not replaced. If it is black, then it is replaced along with one additional ball of the same colour. This process is repeated. Find the probability that the third ball drawn is black.

Answer:

Given urn contains 2 white and 2 black balls

Let us define the events A, B, E_{1}, E_{2}, E_{3}, E_{4}, E are as follows :

A : white ball drawn in 1st turn

B : black ball drawn in 1st turn

E_{1} : white ball drawn in 2nd turn when A has occured

E_{2} : black ball drawn in 2nd turn when A has occured

E_{3} : white ball drawn in 2nd turn when B has occured

E_{4} : black ball drawn in 2nd turn when B has occured

E : black ball drawn in 3rd turn.

Then by law of total prob., we have

P(E) = P(A)[P(E_{1}/A) P(E/AE_{1}) + P(E_{2}/A)P(E/AE_{2})] + P(B)[P(E_{3}/B) P(E/BE_{3}) + P(E_{4}/B) P(E/BE_{4})]

Aliter : Now, we want to find the prob. of drawing a black ball in 3rd turn.

Let W and B denote the drawing of white and black ball in a draw,

required prob. = P (W_{1}W_{2}B_{3}) + P (W_{1}B_{2}B_{3}) + P (Bj W_{2}B_{3}) + P (B_{1}B_{2}B_{3})

Question 11.

Bag A contains 4 white balls and 3 black balls, while bag B contains 3 white balls and 5 black balls. Two balls are drawn from bag a and placed in bag B. Then, what is the probability of drawing a white ball from bag B?

Answer:

Given bag A contains 4 white balls and 3 black balls and bag B contains 3 white balls and 5 black balls.

Let us defme the events are as follows:

E_{1} : event that 2 white balls drawing from bag A and putting to bag B.

E_{2} : drawing 1 white and I black ball from bag A and putting to bag B.

E_{3} : puffing 2 black balls from bag A to bag B

E : drawing a white ball from bag B

Thus E, E_{1}, E_{2}, E_{3} are mutually exclusive and exhaustive events.

P(E/E_{1}) = probability of drawing a white ball from bag B when it is known that 2 white balls are already drawn from bag

A and put into bag B = \(\frac{{ }^5 C_1}{{ }^{10} C_1}=\frac{5}{10}=\frac{1}{2}\)

when E_{2} has occured i.e. 1 white and I black ball drawn from bag A and put into bag B. Then bag B contains 4 white and 6 black balls.

P(E/E_{2}) = \(\frac{4}{10}=\frac{2}{5}\)

when E_{3} has occured i.e. when 2 black balls drawing from bag A has been put into bag B then bag B contains 3 white balls and 7 black balls

P(E/E_{3}) = \(\frac{3}{10}\)

Then by using law of total probability required probability

= P(E) = P(E_{1}) P(E/E_{1}) + P(E_{2}) P(E/E_{2}) + P(E_{3}) P(E/E_{3})

= \(\frac{2}{7} \times \frac{1}{2}+\frac{4}{7} \times \frac{2}{5}+\frac{1}{7} \times \frac{3}{10}=\frac{1}{7}+\frac{8}{35}+\frac{3}{70}\)

= \(\frac{10+16+3}{70}=\frac{29}{70}\)