Question:

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:

Let A = event of first group winning Let B = event of second group winning Let C = event of introducing a new product

P(A) = 0.6 P(B) = 0.4 P(C|A) = 0.7 P(C|B) = 0.3

P(B|C) = (P(B)*P(C|B))/(P(A)*P(C|A) + P(B)*P(C|B))

P(B|C) = (0.40.3)/(0.60.7 + 0.4*0.3)

P(B|C) = 0.2/0.9

P(B|C) = 0.2222

Question:

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 hosteller?

Answer:

Given,

Total students in college = 100% Students residing in hostel = 60% Students not residing in hostel (Day scholars) = 40%

A grade attained by hostellers = 30% A grade attained by day scholars = 20%

To find,

Probability that the student is a hosteller = ?

Solution:

Let P(H) = Probability that the student is a hosteller Let P(D) = Probability that the student is a day scholar

P(H) = (Probability of getting A grade given that the student is a hosteller) x (Probability of being a hosteller)

P(H) = (30/100) x (60/100)

P(H) = 0.18

Therefore, the probability that the student is a hosteller is 0.18.

Question:

If A and B are two events such that A⊂B and P(B)=0, then which of the following is correct ? This question has multiple correct options A P(A∣B)=P(A)P(B)​ B P(A∣B)<P(A) C P(A∣B)≥P(A) D None of these

Answer:

A. P(A∣B)=P(A)P(B) is correct.

Question:

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:

Step 1: Find the total number of balls in the two bags. Total number of balls = 12

Step 2: Find the total number of red balls in the two bags. Total number of red balls = 6

Step 3: Find the probability of selecting the first bag. Number of balls in the first bag = 8 Probability of selecting the first bag = 8/12 = 2/3

Step 4: Find the probability of drawing a red ball from the first bag. Number of red balls in the first bag = 4 Probability of drawing a red ball from the first bag = 4/8 = 1/2

Step 5: Find the probability of drawing a red ball from the first bag given that a red ball was drawn from one of the two bags. Probability = (Probability of selecting the first bag * Probability of drawing a red ball from the first bag) / Probability of drawing a red ball from any of the two bags = (2/3 * 1/2) / (6/12) = 2/6 = 1/3

Question:

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 it was produced by A?

Answer:

Step 1: Calculate the total number of items produced by each operator.

A = 50% of total items produced B = 30% of total items produced C = 20% of total items produced

Step 2: Calculate the total number of defective items produced by each operator.

A = 1% of total items produced by A B = 5% of total items produced by B C = 7% of total items produced by C

Step 3: Calculate the probability that the defective item was produced by A.

Probability = (Number of defective items produced by A) / (Total number of defective items produced)

Probability = (1% of total items produced by A) / (1% + 5% + 7%)

Probability = 1% / 13%

Probability = 0.0769

Question:

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:

Step 1: Calculate the probability of getting 1,2,3 or 4 with the die. There are 6 possible outcomes when throwing a die: 1,2,3,4,5,6. Therefore, the probability of getting 1,2,3 or 4 is 4/6 or 2/3.

Step 2: Calculate the probability of getting exactly one head when tossing a coin once. The probability of getting exactly one head when tossing a coin once is 1/2.

Step 3: Calculate the probability of getting exactly one head when tossing a coin three times. The probability of getting exactly one head when tossing a coin three times is 3/8.

Step 4: Calculate the probability of the girl throwing 1,2,3 or 4 with the die and obtaining exactly one head. The probability of the girl throwing 1,2,3 or 4 with the die and obtaining exactly one head is (2/3) x (1/2) = 1/3.

Question:

There are three coins. One is a two-headed coin 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:

Answer:

1. The probability of the two-headed coin being chosen at random is 1/3.
2. The probability of the two-headed coin showing heads is 100%.
3. Therefore, the probability that the coin that was tossed and showed heads was the two-headed coin is 1/3.

Question:

Probability that A speaks truth is 54​. A coin is tossed. A reports that a head appears. The probability that actually there was head is A 54​ B 21​ C 51​ D 52

Answer:

Answer: A 54

Question:

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:

Answer: Step 1: The first ball drawn from the urn is red with a probability of 5/10 or 0.5.

Step 2: After 2 additional red balls are put in the urn, there are now 7 red balls and 5 black balls.

Step 3: The probability of drawing a red ball from the urn the second time is 7/12 or 0.583.

Question:

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:

Step 1: Total number of insured persons = 2000 + 4000 + 6000 = 12000

Step 2: Total probability of an accident = 0.01 + 0.03 + 0.15 = 0.19

Step 3: Probability of an accident for a scooter driver = 0.01

Step 4: Probability of an accident for a scooter driver out of the total insured persons = 0.01/0.19 = 0.0526

Question:

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 were defective 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:

1. Machine A produced 60% of the items of output and machine B produced 40% of the items.

2. 2% of the items produced by machine A were defective and 1% produced by machine B were defective.

3. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective.

4. The probability that it was produced by machine B is 40% x 1% = 0.4%.

Question:

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:

Answer:

Given: P(knows answer) = 3/4 P(guesses) = 1/4 P(correct | guesses) = 1/4

We are asked to find: P(knows answer | correct)

Using Bayes’ Theorem: P(knows answer | correct) = (P(correct | knows answer) * P(knows answer)) / P(correct)

P(correct) = P(correct | knows answer) * P(knows answer) + P(correct | guesses) * P(guesses)

P(correct) = (1 * 3/4) + (1/4 * 1/4)

P(correct) = 3/4 + 1/16

P(correct) = 13/16

P(knows answer | correct) = (1 * 3/4) / (13/16)

P(knows answer | correct) = 3/13

Question:

A laboratory blood test is 99% effective in detecting a certain disease when it is 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:

Answer:

Step 1: Calculate the probability that a person has the disease given a positive test result:

P(Disease | Positive Test) = P(Positive Test | Disease) * P(Disease) / P(Positive Test)

Step 2: Calculate the probability of a positive test result:

P(Positive Test) = P(Positive Test | Disease) * P(Disease) + P(Positive Test | No Disease) * P(No Disease)

Step 3: Substitute the given values into the equation:

P(Disease | Positive Test) = 0.99 * 0.001 / (0.99 * 0.001 + 0.005 * 0.999)

Step 4: Simplify the equation:

P(Disease | Positive Test) = 0.1647

Question:

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. If the probability of the lost card is a diamond is p enter 100p

Answer:

Step 1: Determine the probability of the lost card being a diamond. This can be found by dividing the number of diamonds in the pack (13 diamonds) by the total number of cards in the pack (52 cards).

Step 2: Calculate the probability by multiplying the fraction by 100. This will give us the probability of the lost card being a diamond in percent.

Step 3: Enter the result into the equation 100p, where p is the probability of the lost card being a diamond.

Answer: 100p = 100 x (13/52) = 25%