Chapter 13 Probability Ex 13.3 Solutions
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?
Answer - 1 : -
Given urn contains 5red and 5 black balls.
Let in first attemptthe ball drawn is of red colour.
⇒ P (probabilityof drawing a red ball) = 5/10 = ½
Now the two balls ofsame colour (red) are added to the urn then the urn contains 7 red and 5 blackballs.
⇒ P (probabilityof drawing a red ball) = 7/12
Now let in firstattempt the ball drawn is of black colour.
⇒ P (probabilityof drawing a black ball) = 5/10 = ½
Now the two balls ofsame colour (black) are added to the urn then the urn contains 5 red and 7black balls.
⇒ P (probabilityof drawing a red ball) = 5/12
Therefore, theprobability of drawing the second ball as of red colour is:
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.
Answer - 2 : -
Let E1 bethe event of choosing the bag I, E2 be the event of choosingthe bag say bag II and A be the event of drawing a red ball.
Then P (E1)= P (E2) = 1/2
Also P (A|E1)= P (drawing a red ball from bag I) = 4/8 = ½
And P (A|E2)= P (drawing a red ball from bag II) = 2/8 = ¼
Now the probability ofdrawing a ball from bag I, being given that it is red, is P (E1|A).
By using Bayes’theorem, we have:
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?
Answer - 3 : -
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?
Answer - 4 : -
Let E1 bethe event that the student knows the answer, E2 be the eventthat the student guess the answer and A be the event that the answer iscorrect.
Then P (E1)= ¾
And P (E2)= ¼
Also P (A|E1)= P (correct answer given that he knows) = 1
And P (A|E2)= P (correct answer given that he guesses) = ¼
Now the probabilitythat he knows the answer, being given that answer is correct, is P (E1|A).
By using Bayes’theorem, we have:
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?
Answer - 5 : -
Let E1 bethe event that person has a disease, E2 be the event thatperson don not have a disease and A be the event that blood test is positive.
As E1 andE2 are the events which are complimentary to each other.
Then P (E1)+ P (E2) = 1
⇒ P (E2)= 1 – P (E1)
Then P (E1)= 0.1% = 0.1/100 = 0.001 and P (E2) = 1 – 0.001 = 0.999
Also P (A|E1)= P (result is positive given that person has disease) = 99% = 0.99
And P (A|E2)= P (result is positive given that person has no disease) = 0.5% = 0.005
Now the probabilitythat person has a disease, give that his test result is positive is P (E1|A).
By using Bayes’theorem, we have
Question - 6 : - There are three coins. One is a 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?
Answer - 6 : -
Let E1 bethe event of choosing a two headed coin, E2 be the event ofchoosing a biased coin and E3 be the event of choosing anunbiased coin. Let A be the event that the coin shows head.
Then P (E1)= P (E2) = P (E3) = 1/3
As we a headed coinhas head on both sides so it will shows head.
Also P (A|E1)= P (correct answer given that he knows) = 1
And P (A|E2)= P (coin shows head given that the coin is biased) = 75% = 75/100 = ¾
And P (A|E3)= P (coin shows head given that the coin is unbiased) = ½
Now the probabilitythat the coin is two headed, being given that it shows head, is P (E1|A).
By using Bayes’theorem, we have
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?
Answer - 7 : -
Let E1 bethe event that the driver is a scooter driver, E2 be the eventthat the driver is a car driver and E3 be the event that thedriver is a truck driver. Let A be the event that the person meet with anaccident.
Total number ofdrivers = 2000 + 4000 + 6000 = 12000
Then P (E1)= 2000/12000 = 1/6
P (E2) =4000/12000 = 1/3
P (E3) =6000/12000 = ½
As we a headed coinhas head on both sides so it will shows head.
Also P (A|E1)= P (accident of a scooter driver) = 0.01 = 1/100
And P (A|E2)= P (accident of a car driver) = 0.03 = 3/100
And P (A|E3)= P (accident of a truck driver) = 0.15 = 15/100 = 3/20
Now the probabilitythat the driver is a scooter driver, being given that he met with an accident,is P (E1|A).
By using Bayes’theorem, we have
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?
Answer - 8 : -
Let E1 bethe event that item is produced by A, E2 be the event that itemis produced by B and X be the event that produced product is found to bedefective.
Then P (E1)= 60% = 60/100 = 3/5
P (E1) =40% = 40/100 = 2/5
Also P (X|E1)= P (item is defective given that it is produced by machine A) = 2% = 2/100 =1/50
And P (X|E2)= P (item is defective given that it is produced by machine B) = 1% = 1/100
Now the probabilitythat item is produced by B, being given that item is defective, is P (E2|A).
By using Bayes’theorem, we have
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.
Answer - 9 : -
Let E1 bethe event that first group wins the competition, E2 be theevent that that second group wins the competition and A be the event ofintroducing a new product.
Then P (E1)= 0.6 and P (E2) = 0.4
Also P (A|E1)= P (introducing a new product given that first group wins) = 0.7
And P (A|E2)= P (introducing a new product given that second group wins) = 0.3
Now the probability ofthat new product introduced was by the second group, being given that a newproduct was introduced, is P (E2|A).
By using Bayes’theorem, we have
P (E2|A) =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?
Answer - 10 : -
let E1 bethe event that the outcome on the die is 5 or 6, E2 be theevent that the outcome on the die is 1, 2, 3 or 4 and A be the event gettingexactly head.
Then P (E1)= 2/6 = 1/3
P (E2) =4/6 = 2/3
As in throwing a cointhree times we get 8 possibilities.
{HHH, HHT, HTH, THH,TTH, THT, HTT, TTT}
⇒ P (A|E1)= P (obtaining exactly one head by tossing the coin three times if she get 5 or6) = 3/8
And P (A|E2)= P (obtaining exactly one head by tossing the coin three times if she get 1,2, 3 or 4) = ½
Now the probabilitythat the girl threw 1, 2, 3 or 4 with a die, being given that she obtainedexactly one head, is P (E2|A).
By using Bayes’theorem, we have
P (E2|A) =8/11