12th NCERT Probability Exercise 13.3 Questions 14
Do or do not
There is no try

Question (1)

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

Solution

There are 5 black and 5 red, totally 10 balls in a urn.
Let A be the event that the ball drawn in second draw is red.
Let E1 be the event that ball drawn in first draw is black.
Let E2 be the event that ball drawn in first draw is red.
$P\left( {{E_1}} \right) = \frac{{5{C_1}}}{{10{C_1}}}$ $= \frac{5}{{10}} = \frac{1}{2}$ $P\left( {{E_2}} \right) = \frac{{5{C_1}}}{{10{C_1}}}$ $= \frac{5}{{10}} = \frac{1}{2}$ $P\left( {A/{E_1}} \right) = \frac{{5{C_1}}}{{12{C_1}}}$ $= \frac{5}{{12}}$ $P\left( {A/{E_2}} \right) = \frac{{7{C_1}}}{{12{C_1}}}$ $= \frac{7}{{12}}$ $P\left( A \right) = P\left( {{E_1}} \right)P\left( {A/{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A/{E_2}} \right)$ $= \frac{1}{2} \times \frac{5}{{12}} + \frac{1}{2} \times \frac{7}{{12}}$ $= \frac{{5 + 7}}{{24}}$ $= \frac{{12}}{{24}}$ $= \frac{{1}}{{2}}$

Question (2)

A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Solution

 Bag I Bag II Red 4 2 Black; 4 6 Total 8 8

Let A be the event the ball drawn is red.
Let E1 be the event the ball is selected from bag I.
Let E2 be the event the ball is selected from bag II.
The bags are selected at random so, $P\left( {{E_1}} \right) = \frac{1}{2}$ $P\left( {{E_2}} \right) = \frac{1}{2}$ $P\left( {A/{E_1}} \right) = \frac{{4{C_1}}}{{8{C_1}}}$ $= \frac{4}{8}$ $P\left( {A/{E_2}} \right) = \frac{{2{C_1}}}{{8{C_1}}}$ $= \frac{2}{8}$ $P\left( A \right) = P\left( {{E_1}} \right)P\left( {A/{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A/{E_2}} \right)$ $= \frac{1}{2} \times \frac{4}{8} + \frac{1}{2} \times \frac{2}{8}$ $= \frac{{4 + 2}}{{16}}$ $= \frac{6}{{16}}$ By Baye's theorem. $P\left( {{E_1}/A} \right) = \frac{{P\left( {{E_1} \cap A} \right)}}{{P\left( A \right)}}$ $= \frac{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right)}}{{P\left( A \right)}}$ $= \frac{{\frac{1}{2} \times \frac{4}{8}}}{{\frac{6}{{16}}}}$ $= \frac{4}{6}$ $= \frac{2}{3}$

Question (3)

Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is hostler?

Solution

Let A be the event that selected students has grade A.
Let E1 be the event that student resides in hostel.
Let E2 be the event that student does not reside in hostel.
$P\left( {{E_1}} \right) = 0.6,P\left( {{E_2}} \right) = 0.4$ $P\left( {A/{E_1}} \right) = 0.3$ $P\left( {A/{E_2}} \right) = 0.2$ By Baye's theorem, $P\left( {{E_1}/A} \right) = \frac{{P\left( {{E_1} \cap A} \right)}}{{P\left( A \right)}}$ $= \frac{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right)}}{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A/{E_2}} \right)}}$ $= \frac{{0.6 \times 0.3}}{{0.6 \times 0.3 + 0.4 \times 0.2}}$ $= \frac{{0.18}}{{0.18 + 0.08}}$ $= \frac{{0.18}}{{0.26}}$ $= \frac{{9}}{{13}}$

Question (4)

In answering a question on a multiple choice test, a student either knows the answer or guesses. Let 3/4 be the probability that he knows the answer and 1/4 be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability 1/4. What is the probability that the student knows the answer given that he answered it correctly?

Solution

Let A be the event that student gives the answer correctly.
Let E1 be the event he knows the answer.
And E2 be the event he guess the answer.
$P\left( {{E_1}} \right) = \frac{3}{4},P\left( {{E_2}} \right) = \frac{1}{4}$ $P\left( {A/{E_1}} \right) = 1$ $P\left( {A/{E_2}} \right) = \frac{1}{4}$ By Baye's theorem, $P\left( {{E_1}/A} \right) = \frac{{P\left( {{E_1} \cap A} \right)}}{{P\left( A \right)}}$ $= \frac{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right)}}{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A/{E_2}} \right)}}$ $= \frac{{\frac{3}{4} \times 1}}{{\frac{3}{4} \times 1 + \frac{1}{4} \times \frac{1}{4}}}$ $= \frac{{\frac{3}{4}}}{{\frac{3}{4} + \frac{1}{{16}}}}$ $= \frac{{\frac{3}{4}}}{{\frac{{12 + 1}}{{16}}}}$ $= \frac{3}{4} \times \frac{{16}}{{13}}$ $= \frac{{12}}{{13}}$

Question (5)

A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?

Solution

Let A be the event that blood test is positive
E1 is the event that person is suffering from the disease
E2is the event that person is not suffering from the disease
$P\left( {{E_1}} \right) = 0.001,P\left( {{E_2}} \right) = 1 - P\left( {{E_1}} \right) = 1 - 0.001 = 0.999$ $P\left( {A/{E_1}} \right) = 0.99,P\left( {A/{E_2}} \right) = 0.005,$ $P\left( {{E_1}/A} \right) = \frac{{P\left( {{E_1} \cap A} \right)}}{{P\left( A \right)}}$ $= \frac{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right)}}{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A/{E_2}} \right)}}$ $= \frac{{0.001 \times 0.99}}{{0.001 \times 0.99 + 0.999 \times 0.005}}$ $= \frac{{0.00099}}{{0.00099 + 0.004995}}$ $= \frac{{990}}{{5985}}$ $= \frac{{22}}{{133}}$

Question (6)

There are three coins. One is two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?

Solution

Let A be the event that coin shows head.
E1 be the event that coin is two headed coin.
E2 be the event that coin is a baisedd coin.
E3 be the event that coin is an unbaised coin.
As the coin is selected at random. $P\left( {{E_1}} \right) = \frac{1}{3},P\left( {{E_2}} \right) = \frac{1}{3},P\left( {{E_3}} \right) = \frac{1}{3}$ $P\left( {A/{E_1}} \right) = 1,P\left( {A/{E_2}} \right) = 0.75 = \frac{3}{4},P\left( {A/{E_3}} \right) = \frac{1}{2}$ $P\left( {{E_1}/A} \right) = \frac{{P\left( {{E_1} \cap A} \right)}}{{P\left( A \right)}}$ $= \frac{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right)}}{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A/{E_2}} \right) + P\left( {{E_3}} \right)P\left( {A/{E_3}} \right)}}$ $= \frac{{\frac{1}{3} \times 1}}{{\frac{1}{3} \times 1 + \frac{1}{3} \times \frac{3}{4} + \frac{1}{3} \times \frac{1}{2}}}$ $= \frac{{\frac{1}{3}}}{{\frac{1}{3} + \frac{3}{{12}} + \frac{1}{6}}}$ $= \frac{1}{3} \times \frac{{12}}{9}$ $= \frac{4}{9}$

Question (7)

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

Solution

Let A be the event that insured person meets with an accident.
Let E1 be the event that it is scooter driver.
Let E2 be the event that it is car driver.
Let E3 be the event that it is truck driver.
$P\left( {{E_1}} \right) = \frac{{2000}}{{12000}} = \frac{2}{{12}},P\left( {{E_2}} \right) = \frac{{4000}}{{12000}} = \frac{4}{{12}},P\left( {{E_3}} \right) = \frac{{6000}}{{12000}} = \frac{6}{{12}}$ $P\left( {A/{E_1}} \right) = 0.01 = \frac{1}{{100}},P\left( {A/{E_2}} \right) = 0.03 = \frac{3}{{100}},P\left( {A/{E_3}} \right) = 0.15 = \frac{{15}}{{100}}$ $P\left( {{E_1}/A} \right) = \frac{{P\left( {{E_1} \cap A} \right)}}{{P\left( A \right)}}$ $= \frac{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right)}}{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A/{E_2}} \right) + P\left( {{E_3}} \right)P\left( {A/{E_3}} \right)}}$ $= \frac{{\frac{2}{{12}} \times \frac{1}{{100}}}}{{\frac{2}{{12}} \times \frac{1}{{100}} + \frac{4}{{12}} \times \frac{3}{{100}} + \frac{4}{{12}} \times \frac{{15}}{{100}}}}$ $= \frac{{\frac{2}{{1200}}}}{{\frac{{2 + 12 + 90}}{{1200}}}}$ $= \frac{2}{{104}}$ $= \frac{1}{{52}}$

Question (8)

A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that was produced by machine B?

Solution

Let A be the event that the item selected is defective.
Let E1 is the event that selected items is from Machine A.
Let E2 is the event that selected items is from Machine B.
$P\left( {{E_1}} \right) = \frac{{60}}{{100}} = \frac{3}{5},P\left( {{E_2}} \right) = \frac{{40}}{{100}} = \frac{2}{5}$ $P\left( {A/{E_1}} \right) = \frac{2}{{100}},P\left( {A/{E_2}} \right) = \frac{1}{{100}}$ $P\left( {{E_2}/A} \right) = \frac{{P\left( {{E_2} \cap A} \right)}}{{P\left( A \right)}}$ $= \frac{{P\left( {{E_2}} \right)P\left( {A/{E_2}} \right)}}{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A/{E_2}} \right)}}$ $= \frac{{\frac{2}{5} \times \frac{1}{{100}}}}{{\frac{3}{5} \times \frac{2}{{100}} + \frac{2}{5} \times \frac{1}{{100}}}}$ $\frac{{\frac{2}{{500}}}}{{\frac{{6 + 2}}{{500}}}}$ $= \frac{2}{8}$ $= \frac{1}{4}$

Question (9)

Two groups are competing for the position on the board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group

Solution

Let A be the event that winning group will intoduced new product.
Let E1 be the event that first group will win .
Let E1 be the event that first group will win .
$P\left( {{E_1}} \right) = \frac{6}{{10}} = \frac{3}{5},P\left( {{E_2}} \right) = \frac{4}{{10}} = \frac{2}{5}$ $P\left( {A/{E_1}} \right) = \frac{7}{{10}},P\left( {A/{E_2}} \right) = \frac{3}{{10}}$ $P\left( {{E_2}/A} \right) = \frac{{P\left( {{E_2} \cap A} \right)}}{{P\left( A \right)}}$ $= \frac{{P\left( {{E_2}} \right)P\left( {A/{E_2}} \right)}}{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A/{E_2}} \right)}}$ $= \frac{{\frac{2}{5} \times \frac{3}{{10}}}}{{\frac{3}{5} \times \frac{7}{{10}} + \frac{2}{5} \times \frac{3}{{10}}}}$ $= \frac{{\frac{6}{{50}}}}{{\frac{{21 + 6}}{{50}}}}$ $= \frac{6}{{27}}$ $= \frac{2}{{9}}$

Question (10)

Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?

Solution

Let A be the event that head appears on the coin.
Let E1 be the event that number 5 or 6 appears on the die.
Let E2 be the event that number1, 2, 3 and 4 appears on the die.
$P\left( {{E_1}} \right) = \frac{2}{6} = \frac{1}{3},P\left( {{E_2}} \right) = \frac{4}{6} = \frac{2}{3}$ $P\left( {A/{E_1}} \right) = \frac{3}{8},P\left( {A/{E_2}} \right) = \frac{4}{8}$ $P\left( {{E_2}/A} \right) = \frac{{P\left( {{E_2} \cap A} \right)}}{{P\left( A \right)}}$ $= \frac{{P\left( {{E_2}} \right)P\left( {A/{E_2}} \right)}}{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A/{E_2}} \right)}}$ $\frac{{\frac{2}{3} \times \frac{4}{8}}}{{\frac{1}{3} \times \frac{3}{8} + \frac{2}{3} \times \frac{4}{8}}}$ $= \frac{{\frac{8}{{24}}}}{{\frac{{3 + 8}}{{24}}}}$ $= \frac{8}{{11}}$

Question (11)

A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that was produced by A?

Solution

Let A be the event that defective item is produced.
Let E1 be the event that it has been produced by operater A.
Let E2 be the event that it has been produced by operater B.
Let E3 be the event that it has been produced by operater C.
$P\left( {{E_1}} \right) = \frac{{50}}{{100}} = \frac{5}{{10}},P\left( {{E_2}} \right) = \frac{{30}}{{100}} = \frac{3}{{10}},P\left( {{E_3}} \right) = \frac{{20}}{{100}} = \frac{2}{{10}}$ $P\left( {A/{E_1}} \right) = \frac{1}{{100}},P\left( {A/{E_2}} \right) = \frac{5}{{100}},P\left( {A/{E_3}} \right) = \frac{7}{{100}}$ $P\left( {{E_1}/A} \right) = \frac{{P\left( {{E_1} \cap A} \right)}}{{P\left( A \right)}}$ $= \frac{{P\left( {{E_1}} \right)P\left( {A/{E_2}} \right)}}{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A/{E_2}} \right) + P\left( {{E_2}} \right)P\left( {A/{E_2}} \right)}}$ $= \frac{{\frac{5}{{10}} \times \frac{1}{{100}}}}{{\frac{5}{{10}} \times \frac{1}{{100}} + \frac{3}{{10}} \times \frac{5}{{100}} + \frac{2}{{10}} \times \frac{7}{{100}}}}$ $= \frac{{\frac{5}{{1000}}}}{{\frac{{5 + 15 + 14}}{{1000}}}}$ $= \frac{5}{{34}}$

Question (12)

A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

Solution

Let A be the event that both the cards drawn are diamonds.
Let E1 be the event that diamond card is lost.
Let E2 be the event that non diamond card is lost.
$P\left( {{E_1}} \right) = \frac{{13}}{{52}} = \frac{1}{4},P\left( {{E_2}} \right) = \frac{{39}}{{52}} = \frac{3}{4}$ $P\left( {A/{E_1}} \right) = \frac{{12{C_2}}}{{51{C_2}}} = \frac{{12 \times 11}}{{51 \times 50}} = \frac{{22}}{{425}}$ $P\left( {A/{E_2}} \right) = \frac{{13{C_2}}}{{51{C_2}}} = \frac{{13 \times 12}}{{51 \times 50}} = \frac{{26}}{{425}}$ By Baye's theorem, $P\left( {{E_1}/A} \right) = \frac{{P\left( {{E_1} \cap A} \right)}}{{P\left( A \right)}}$ $= \frac{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right)}}{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A/{E_2}} \right)}}$ $= \frac{{\frac{1}{4} \times \frac{{22}}{{425}}}}{{\frac{1}{4} \times \frac{{22}}{{425}} + \frac{3}{4} \times \frac{{26}}{{425}}}}$ $= \frac{{\frac{1}{4} \times \frac{{22}}{{425}}}}{{\frac{{22 + 78}}{{4 \times 425}}}}$ $= \frac{{22}}{{100}}$ $= \frac{{11}}{{50}}$

Question (13)

Probability that A speaks truth is 4/5. A coin is tossed. A reports that a head appears. The probability that actually there was head is
A. 4/5 B. 1/2 C. 1/5 D. 2/5

Solution

Let A be the event that a man reports that head appears on coin.
E1 is event that head appears on coin.
E2 is event that head does not appear on the coin.
$P\left( {{E_1}} \right) = \frac{1}{2},P\left( {{E_2}} \right) = \frac{1}{2}$ $P\left( {A/{E_1}} \right) = \frac{4}{5},P\left( {A/{E_2}} \right) = \frac{1}{5}$ $P\left( {{E_1}/A} \right) = \frac{{P\left( {{E_1} \cap A} \right)}}{{P\left( A \right)}}$ $= \frac{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right)}}{{P\left( {{E_1}} \right)P\left( {A/{E_1}} \right) + P\left( {{E_2}} \right)P\left( {A/{E_2}} \right)}}$ $= \frac{{\frac{1}{2} \times \frac{4}{5}}}{{\frac{1}{2} \times \frac{4}{5} + \frac{1}{2} \times \frac{1}{5}}}$ $= \frac{{\frac{1}{2} \times \frac{4}{5}}}{{\frac{{4 + 1}}{{2 \times 5}}}}$ $= \frac{4}{5}$ So A is the correct answer.

Question (14)

If A and B are two events such that A ⊂ B and P(B) ≠ 0, then which of the following is correct?
A. P(A|B) = P(B)/ P(A) B. P(A|B) < P(A) C. P(A|B) ≥ P(A) D. None of these

Solution

If A and B are two events such that A ⊂ B and P(B) ≠ 0,
Since A ⊂ B , A ∩ B = A,
P(A) < P(B) , ⇒ P(A)/ P(B) < 1
$P\left( {A/B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}$ $= \frac{{P\left( A \right)}}{{P\left( B \right)}}$ P(A|B) ≥ P(A)
So C is the corrrect answer.