2 % of the population have a certain blood disease of a serious form: 10 % have it in a mild form; and 88 % don't have it at all. A new blood test is developed; the probability of testing positive is 9 10 if the subject has the serious form, 6 10 if the subject has the mild form, and 1 10 if the subject doesn't have the disease. A subject is tested positive. What is the probability that the subject has serious form of the disease?

Let the person that has disease in serious form = A 1 Let the person that has disease in mild form = A 2 Let the person that doesn't have a disease = A 3 Let the test positive = B Thus, P ( A 1 ) = 0 . 02 , P ( A 2 ) = 0 . 1 , P ( A 3 ) = 0 . 88 P ( B / A 1 ) = 0 . 9 , P ( B / A 2 ) = 0 . 6 , P ( B / A 3 ) = 0 . 1 Now, according to Bayes' Theorem: P ( A 1 / B ) = P ( B / A 1 ) P ( A 1 ) P ( B / A 1 ) P ( A 1 ) + P ( B / A 2 ) P ( A 2 ) + P ( B / A 3 ) P ( A 3 ) = 0 . 9 × 0 . 02 0 . 9 × 0 . 02 + 0 . 1 × 0 . 6 + 0 . 88 × 0 . 1 = 0 . 108 Thus, the probability of having a serious form of disease is 0 . 108

A doctor is to visit a patient. From the past experience, it is known that the probabilities that he will come by train, bus, scooter or by other means of transport are respectively 3 10 , 1 5 , 1 10 and 2 5 The Probabilities that he will be late are 1 4 , 1 3 , 1 12 , if he comes by train, bus and scooter respectively, but if he comes by other means of transport, he will not be late. When he arrives, he is late. What is the probability that he comes by train?

Given that a doctor is to visit a patient. The probability that he comes by train, bus, scooter and other transport are 3 10 , 1 5 , 1 10 and 2 5 respectively. Probability of doctor arriving late if he comes by train, bus and scooter are 1 4 , 1 3 and 1 12 respectively. Let E is an event that doctor will be late to visit patient. Event that doctor comes by train, bus, scooter and other transport be denoted by T 1 , T 2 , T 3 and T 4 respectively. Then P ( T 1 ) = 3 10 , P ( T 2 ) = 1 5 , P ( T 3 ) = 1 10 and P ( T 4 ) = 2 5 ∴ Probability that doctor will be late given that he comes by train is P ( E T 1 ) = 1 4 , Probability that doctor will be late given that he comes by bus is P ( E T 2 ) = 1 3 and Probability that doctor will be late given that he comes by scooter is P ( E T 3 ) = 1 12 Probability that doctor will be late given that he comes by other means if transport is P ( E T 4 ) = 0 Using Bayes' theorem, Probability that doctor comes by train given than he is late = P ( T 1 E ) = P ( T 1 ) · P ( E T 1 ) P ( T 1 ) · P ( E T 1 ) + P ( T 2 ) · P ( E T 2 ) + P ( T 3 ) · P ( E T 3 ) + P ( T 4 ) · P ( E T 4 ) = 3 10 × 1 4 3 10 × 1 4 + 1 5 × 1 3 + 1 10 × 1 12 + 2 5 × 0 = 3 40 3 40 + 1 15 + 1 120 + 0 = 3 × 120 40 × 8 = 1 2 Hence, the required probability is 1 2 .

Ten cards numbered 1 to 10 are placed in box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3 , what is the probability that it is an even number ?

Ten cards numbered 1 to 10 are placed in a box and one card is drawn randomly. To find: Probability that number on card drawn is even given that the number is more than 3 Let A : event that number on card drawn is even number and B : event that number on card drawn is more than 3 . Sample space S = 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 . Then, A = 2 , 4 , 6 , 8 , 10 B = 4 , 5 , 6 , 7 , 8 , 9 , 10 So, A ∩ B = 4 , 6 , 8 , 10 Thus, we get P A = 5 10 , P B = 7 10 and P A ∩ B = 4 10 We know that P A | B = P A ∩ B P B ⇒ P ( A | B ) = 4 / 10 7 / 10 = 4 7 which is the required probability.

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?

Consider E 1 and E 2 be the events that the student is a hostler and a day scholar respectively and A be the event that the chosen student gets grade A . ∴ P E 1 = 60 % = 60 100 = 0.6 P E 2 = 40 % = 40 100 = 0.4 P A | E 1 = P (student getting an A grade is a hostler) = 30 % = 0.3 P A | E 2 = P (student getting an A grade is a day scholar) = 20 % = 0.2 So, the probability that a randomly chosen student is a hostler, given that he has an A grade, is given by P E 1 A . By using Bayes’ theorem, we get: P ( E 1 | A ) = P E 1 × P A E 1 P E 1 × P A E 1 + P E 2 × P A E 2 = 0.6 × 0.3 0.6 × 0.3 + 0.4 × 0.2 = 0.18 0.26 = 9 13 Hence, the required probability is 9 13

A family has two children. What is the probability that both children are boys, given that at least one of them is a boy?

Given, a family has two children. Let A be the event that both children are boys B be the event that at least one of them is a boy. Then, A = B B and B = B B , B G , G B A ∩ B = B B From the given data, P A = 1 4 P B = 3 4 and P A ∩ B = 1 4 The required probability is P A B = P A ∩ B P B = 1 / 4 3 / 4 P A / B = 1 3 Therefore, the required probability is 1 3 .

A bag contains 5 white, 7 red and 4 black balls. If four balls are drawn one by one with replacement, what is the probability that none is white?

Given that, a bag contains 5 white, 7 red and 4 black balls. Here, total number of balls in the bag = 5 + 7 + 4 = 16 Probability of getting a white ball = 5 16 Probability of not getting white ball = 11 16 Four balls are drawn one by one with replacement Therefore, Probability that none is white = 11 16 × 11 16 × 11 16 × 11 16 = 11 16 4 = 14641 65536 Hence, the probability of that none is white = 14641 65536 .

The persons A , B and C shoot to hit a target. Their probabilities of hitting the target are 5 6 , 4 5 and 3 4 respectively. Find the probability that at least one person hits the target.

Here, the probabilities of hitting the target by three persons A , B and C are 5 6 , 4 5 and 3 4 respectively. i.e. P ( A ) = 5 6 , P ( B ) = 4 5 , P ( C ) = 3 4 . So, the required probability i.e. at least one person hits the target is = 1 - P A ¯ ∩ B ¯ ∩ C ¯ = 1 - P A ¯ · P B ¯ · P C ¯ = 1 − 1 − 5 6 1 − 4 5 1 − 3 4 = 1 − 1 6 × 1 5 × 1 4 = 1 − 1 120 = 119 120 = 0.991

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 ?

Let E 1 , E 2 , E 3 and A be the events defined E 1 = Two white balls is transferred from bag A to bag B E 2 = Two black balls is transferred from bag A to bag B E 3 = One white and one black balls is transferred from bag A to bag B A = One ball drawn from the bag is white. P ( E 1 ) = 4 C 2 7 C 2 = 4 × 3 2 × 1 7 × 6 2 × 1 = 2 7 [Using C r n = n ! r ! ( n - r ) ! ] P ( E 2 ) = 3 C 2 7 C 2 = 3 × 2 2 × 1 7 × 6 2 × 1 = 1 7 P ( E 3 ) = 4 C 1 × 3 C 2 7 C 2 = 4 × 3 7 × 6 = 4 7 When E 1 has occurred, two white balls has already been transferred from bag A to bag B contains 5 white balls and 5 black balls. P A E 1 = 5 10 When E 2 has occurred two black balls has already been transferred from bag A to bag B contains 3 white balls and 7 black balls. P A E 2 = 3 10 When E 3 has occurred one black ball and one white balls has already been transferred from bag A to bag B contain 4 white balls and 6 black balls. P A E 3 = 4 10 By using law of total probability, P ( A ) = P ( E 1 ) P A E 1 + P ( E 2 ) P A E 2 + P ( E 3 ) P A E 3 = 2 7 5 10 + 1 7 3 10 + 4 7 4 10 = 10 70 + 3 70 + 16 70 = 29 70 Hence, the required probability is 29 70 .

If A and B are the events such that P A = 1 2 , P B = 1 3 and P A ∩ B = 1 4 , then find P B A .

Given that A and B are the events such that P A = 1 2 , P B = 1 3 and P A ∩ B = 1 4 To find P B A : We know that P B A = P A ∩ B P A ⇒ P B A = 1 4 1 2 ⇒ P B A = 1 2 .

If A and B are the events such that P A = 1 2 , P B = 1 3 and P A ∩ B = 1 4 , then find P A B .

Given that A and B are the events such that P A = 1 2 , P B = 1 3 and P A ∩ B = 1 4 To find P A B : We know that P A B = P A ∩ B P B ⇒ P A B = 1 4 1 3 ⇒ P A B = 3 4 .

EASY

12th ICSE

IMPORTANT

Earn 100

State the theorem of total probability.

Important Questions on Probability

MEDIUM

12th ICSE

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Probability>Exercise>Q 96

E_{1}

and

E_{2}

are equally likely, mutually exclusive and exhaustive events and

P (A / E_{1}) = 0.2, P (A / E_{2}) = 0.3

Find

P (E_{1} / A)

HARD

12th ICSE

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Probability>Exercise>Q 97

A doctor is called to see a sick child. The doctor has prior information that

80 %

of the sick children in that area have the flu, while the other

20 %

are sick with measles. Assume that there is no other disease in that area. A well-known symptom of measles is rash. From the past records, it is known that, chances of having rashes given that sick child is suffering from measles is

0.95

. However occasionally children with flu also develop rash, whose chance are

0.08

. Upon examining the child, the doctor finds a rash. What is the probability that child is suffering from measles?

HARD

12th ICSE

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Probability>Exercise>Q 98

2 %

of the population have a certain blood disease of a serious form:

10 %

have it in a mild form; and

88 %

don't have it at all. A new blood test is developed; the probability of testing positive is

\frac{9}{10}

if the subject has the serious form,

\frac{6}{10}

if the subject has the mild form, and

\frac{1}{10}

if the subject doesn't have the disease. A subject is tested positive. What is the probability that the subject has serious form of the disease?

EASY

12th ICSE

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Probability>Exercise>Q 99

A firm manufactures PVC pipes in three plants viz,

X

Y

and

Z

. The daily production volumes from the three firms

X

Y

and

Z

are respectively

2000

units,

3000

units and

5000

units. It is known from the past experience that

3 %

of the output from plant

X

4 %

from plant

Y

and

2 %

from plant

Z

are defective. A pipe is selected at random from a day’s total production, if the selected pipe is a defective, then what is the probability that it was produced by plant

Y

HARD

12th ICSE

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Probability>Exercise>Q 100

The chances of

A

B

and

C

becoming manager of a certain company are

5 : 3 : 2

. The probabilities that the office canteen will be improved if

A

B

, and

C

become managers are

0.4

0.5

and

0.3

respectively. If the office canteen has been improved, what is the probability that

B

was appointed as the manager?

HARD

12th ICSE

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Probability>Exercise>Q 101

Bag

I

contain

3

red and

4

black balls while another Bag

II

contains

5

red and

6

black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag

II

HARD

12th ICSE

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Probability>Exercise>Q 102

A doctor is to visit a patient. From the past experience, it is known that the probabilities that he will come by train, bus, scooter or by other means of transport are respectively $\frac{3}{10}, \frac{1}{5}, \frac{1}{10}$ and $\frac{2}{5}$ The Probabilities that he will be late are $\frac{1}{4}, \frac{1}{3}$ , $\frac{1}{12}$ , if he comes by train, bus and scooter respectively, but if he comes by other means of transport, he will not be late. When he arrives, he is late. What is the probability that he comes by train?

MEDIUM

12th ICSE

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 3 - Probability>Exercise>Q 103

Bag I contains

3

red and

4

black balls while Bag II contains

4

red and

5

black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II and it was found to be red. Find the probability that the transferred ball is black.

State the theorem of total probability.

Important Questions on Probability

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