PROBLEM & SOLUTION -
Example.1) A coin is tossed once. Find the probability of getting tail.
Ans.) When a coin is tossed, we get either Head (H) or Tail (T)
Sample space, S = {T, H}
=> n(S) = 2
Let, E be the event of getting a tail.
Then, E = {T}
=> n(E) = 1
n(E) 1
So, P (getting a tail) = -------- = ----- (Ans.)
n(S) 2
Example.2) Two coins are tossed once. Find the probability of getting (i) 2 heads, and (ii) at least 1 tail.
Ans.) When two coins are tossed once, the sample space is given by
S = {HH, HT, TH, TT}
=> n(S) = 4
(i) Let E₁ be the event of getting two heads.
Then, E₁ = {HH}
=> n(E₁) = 1
n(E₁) 1
So, P(getting 2 heads) = P(E₁) = --------- = ------ (Ans.)
n(S) 4
(ii) Let, E₂ be the event of getting at least one tail. Then
E₂ = {HT, TH, TT}
=> n(E₂) = 3
n(E₂) 3
P(getting at least 1 tail) = P(E₂) = --------- = ------- (Ans.)
n(S) 4
Example.3) A die is thrown once. What is the probability of getting (i) an even number, (ii) a number greater than 2 ?
Ans.) When a die is thrown once, the sample space is given by
S = {1, 2, 3, 4, 5, 6}
=> n(S) = 6.
(i) Let E₁ be the event of getting an even number.
Then, E₁ = {2, 4, 6}
=> n(E₁) = 3
P(getting an even number)
n(E₁) 3 1
=> P(E₁) = --------- = ------- = -------
n(S) 6 2
(ii) Let E₂ be the event of getting a number greater than 2.
Then, E₂ = {3, 4, 5, 6}
=> n(E₂) = 4
So, P (getting a number greater than 2)
n(E₂) 4 2
= P(E₂) = --------- = -------- = ------- (Ans.)
n(S) 6 3
Example.4) Cards marked with numbers 1, 2, 3,…………..,20 are well shuffled and a card is drawn as random. What is the probability that the number on the card is
(i) A prime number ?
(ii) A number divisible by 3 ?
(iii) A perfect square number ?
Ans.) Clearly, the sample space is given by –
S = {1, 2, 3, 4,………………., 20}
(i) Let, E₁ be the event of getting a prime number
Then, E₁ = {2, 3, 5, 7, 11, 13, 17, 19}
=> n(E₁) = 8
So, P(getting a prime number)
n(E₁) 8 2
= P(E₁) = --------- = ------ = ------ (Ans.)
n(S) 20 5
(ii) Let, E₂ be the event of getting a number divisible by 3
Then, E₂ = {3, 6, 9, 12, 15, 18}
=> n(E₂) = 6
So, P(getting a number divisible by 3) =
n(E₂) 6 3
P(E₂) = --------- = ------- = -------- (Ans.)
n(S) 20 10
(iii) Let, E₃ be the event of getting a perfect square number.
Then, E₃ = {1, 4, 9, 16}
=> n(E₃) = 4
So, P(putting a perfect square number) =
n(E₃) 4 1
P(E₃) = ---------- = -------- = ------- (Ans.)
n(S) 20 5
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