## Probability

Q1: A fair coin is tossed twice, find the probability of getting different outcomes of this experiment. What will be the sum of all these probabilities?

Answer: A coin is tossed twice, the sample space (possible outcomes) will be:

S = {HH, HT, TH, TT}

Let A be the event getting two heads then

 Number of Favourable Outcomes for event A 1 P(A) = = Total Number of Outcomes 4

Let B be the event getting first head and then tail (HT) then

 Number of Favourable Outcomes for event B 1 P(B) = = Total Number of Outcomes 4

Let C be the event getting first tail then head (TH) then

 Number of Favourable Outcomes for event C 1 P(C) = = Total Number of Outcomes 4

Let D be the event getting two tails (TT) then

 Number of Favourable Outcomes for event D 1 P(D) = = Total Number of Outcomes 4

Sum of all probabilities P(A) + P(B) + P(C) + P(D) = 4/4 = 1

Q2: Find the probability that a leap year selected randomly will have 53 Sundays?
(Note: If first two days lands on a Sunday in a leap year it would have 53 Sundays. How to prove?).

Answer: No. of days in a leap year = 366 days = 52 weeks + 2 days

It implies a leap year will have 52 Sundays.
In remaining 2 days, possible out comes are:
- Sun, Mon
- Mon, Tue
- Tue, Wed
- Wed, Thu
- Thu, Fri
- Fri, Sat
- Sat, Sun

Total out comes = 7
Favourable outcomes that Sunday will come in these two days = 2

∴ Required probability = 2/7

Q3: A shop keeper has a box containing 10 dolls, out which three are defective. Radha buys a toy from the shop keeper. What is the probability that Radha gets

(i) a defective doll.
(ii) a non-defective doll.

Defectives = 3
Non-Defective = 7

(i) Probability Radha gets a defective doll
 Number of Favourable Outcomes (defective dolls) 3 P(A) = = Total Number of Outcomes 10

P(A) = 0.3

(ii) Probability Radha gets non-defctive doll
 Number of Favourable Outcomes (non-defective dolls) 7 P(B) = = Total Number of Outcomes 10

P(B) = 0.7

Q4: A pair of fair dice are rolled. Find the probability that the two dice show the same number.
Answer: Total possible outcomes when two dice are thrown = 36 = {(1,1), (1,2), ..., (6,5), (6,6)}

Favourable outcomes (two dice show same number) = 6 = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

 Number of Favourable Outcomes (two dice show same number) 6 P(E) = = Total Number of Outcomes 36

P(E) = 1/6

Q5: In the above example, what will be the probability of getting two different numbers when the two dice are thrown?

Answer: As computed above, P(A) = getting same numbers on the two dice = 1/6.
∴ Probability Getting two different numbers P(B) = P (NOT A) = 1 - P(A) = 1 - 1/6 = 5/6  