# NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.3

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## NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.3

Ex 13.3 Class 12 Maths Question 1.
An urn contain 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:
Urn contain 5 red and 5 black balls.
(i) Let a red ball is drawn.
probability of drawing a red ball = $frac { 5 }{ 10 }$ = $frac { 1 }{ 2 }$
Now two red balls are added to the urn.
=> The urn contains 7 red and 5 black balls.
Probability of drawing a red ball = $frac { 7 }{ 12 }$
(ii) Let a black ball is drawn at first attempt
Probability of drawing a black ball = $frac { 5 }{ 10 }$ = $frac { 1 }{ 2 }$
Next two black balls are added to the urn
Now urn contains 5 red and 7 black balls
Probability of getting a red ball = $frac { 5 }{ 12 }$
=> Probability of drawing a second ball as red
$=frac { 1 }{ 2 } times frac { 7 }{ 12 } +frac { 1 }{ 2 } times frac { 5 }{ 12 } =frac { 7 }{ 24 } +frac { 5 }{ 24 } =frac { 12 }{ 24 } =frac { 1 }{ 2 }$

Ex 13.3 Class 12 Maths 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:
Let A be the event that ball drawn is red and let E1 and E2 be the events that the ball drawn is from the first bag and second bag
respectively. P(E1) = $frac { 1 }{ 2 }$, P(E2) = $frac { 1 }{ 2 }$.
P (A|E 1) = Probability of drawing a red ball from bag

Ex 13.3 Class 12 Maths 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 a hostlier?
Solution:
Let E1, E2 and A represents the following:
E1 = students residing in the hostel,
E2 day scholars (not residing in the hostel)
and A = students who attain grade A

Ex 13.3 Class 12 Maths Question 4.
In answering a question on a multiple choice test, a student either knows the answer or 3 guesses. Let $frac { 3 }{ 4 }$ be the probability that he knows the answer and $frac { 1 }{ 4 }$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $frac { 1 }{ 4 }$ . What is the probability that the student knows the answer given that he answered it correctly?
Solution:
Let the event E1 = student knows the answer , E2 = He gusses the answer
P(E1) = $frac { 3 }{ 4 }$,P(E2) = $frac { 1 }{ 4 }$
Let A is the event that answer is correct, if the student knows the answer

Ex 13.3 Class 12 Maths 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
E1 = The person selected is suffering from certain disease,
E2 = The person selected is not suffering from certain disease.
A = The doctor diagnoses correctly

Ex 13.3 Class 12 Maths Question 6.
There are three coins. One is a two headed coin, another is a biased coin that conies up heads 75% of the time and third is an unbiased coin. One of the three coins is choosen at random and tossed, it shows head, what is the probability that it was the two headed coin?
Solution:
Let E1, E2, E3 and A denotes the following:
E1 = a two headed coin, E2 = a biased coin,
E3 = an unbiased coin, A=A head is shown

Ex 13.3 Class 12 Maths Question 7.
An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of mi accident are 0.01, 0.03, 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Solution:
An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. Total number of drivers
=2000 + 4000 + 6000 = 12,000
Probability of selecting a scooter driver

Ex 13.3 Class 12 Maths 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 it was produced by machine B.?
Solution:
E1 and E2 are the events the percentage of production of items by machine A and machine B respectively.
Let A denotes defective item.
Machine A’s production of items = 60 %
Probability of production of items by machine

Ex 13.3 Class 12 Maths 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:
Given: P (G1) = 0.6, P (G2) = 0.4
P represents the launching of new product P(P|G1) = 0.7 and P(P|G2) = 0.3

Ex 13.3 Class 12 Maths 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:
When a die is thrown there are 6 exhaustive cases.
If she gets 5 or 6 the probability of E1 = $frac { 2 }{ 6 }$ = $frac { 1 }{ 3 }$
i.e.,P(E1) = $frac { 1 }{ 3 }$
If she gets 1,2,3,4 and probability of E2 = $frac { 4 }{ 6 }$ = $frac { 2 }{ 3 }$

Ex 13.3 Class 12 Maths Question 11.
A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as 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 it was produced by A?
Solution:
Let E1, E2, E3 and A be the events defined as follows:
E1 = the item is manufactured by the operator A
E2 = the item is manufactured by the operator B
E3 = the item is manufactured by the operator C
and A=the item is defective

Ex 13.3 Class 12 Maths 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:
E1 = Event that lost card is diamond,
E2 = Event that lost card is not diamond.
There are 13 diamond cards, out of a pack or 52 cards

Ex 13.3 Class 12 Maths 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) $frac { 4 }{ 5 }$
(b) $frac { 1 }{ 2 }$
(c) $frac { 1 }{ 5 }$
(d) $frac { 2 }{ 5 }$
Solution:
Let A be the event that the man reports that head occurs in tossing a coin and let E1 be the event that head occurs and E2 be the event head does not occurs.

Ex 13.3 Class 12 Maths 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) = $frac { P(B) }{ P(A) }$
(b) P (A | B) < P (A)
(c) P(A | B) ≥ P(A)
(d) None of these
Solution: (c) A⊂B => A∩B = A and P(B)≠0
$P(A|B)=frac { P(Acap B) }{ P(B) } =frac { P(A) }{ P(B) }$

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