CBSE Class 10 - Maths - Probability - 5 Questions That Your Teacher Can Ask To Confuse You

Probability

5 Questions Your Maths Teacher Can Ask To Confuse You

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. 


Answer: Total Dolls = 10
               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