# CLASS-10PROBABILITY - PROBLEM & SOLUTION

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