Explanation :

Here, n(S) = 6 x 6 = 36

Let E = event of getting a total more than 7.

Then, E = {(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)}

∴ n(E) = 15

∴ P(E) = $ \dfrac{\text{n(E)}}{\text{n(S)}}$

⇒ P(E) = $\dfrac{15}{36}$

⇒ P(E) = $\dfrac{5}{12}$

Explanation :

Here, n(S) = 6 x 6 = 36

Let E = that the sum of the numbers on the two faces is divisible by 4 and 6.

Then, E = {(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2),(6,6)}

∴ n(E) = 14

∴ P(E) = $ \dfrac{\text{n(E)}}{\text{n(S)}}$

⇒ P(E) = $\dfrac{14}{36}$

⇒ P(E) = $\dfrac{7}{18}$

Explanation :

METHOD 1:

In tossing two coins, the sample space is given by,

S = {HH, HT, TH, TT}

∴ n(S) = 4

Let E = event of getting at least one head.

Then, E = {HH,HT,TH}

∴ n(E) = 3

∴ P(E) = $ \dfrac{\text{n(E)}}{\text{n(S)}}$

⇒ P(E) = $\dfrac{3}{4}$

METHOD 2:

If n unbaised coins are tossed. Total number of outcomes in the sample space = 2ⁿ

∴ n(S) = 2² = 4

Let E = event of getting at least one head.

Then, E = {HH,HT,TH}

∴ n(E) = 3

∴ P(E) = $ \dfrac{\text{n(E)}}{\text{n(S)}}$

⇒ P(E) = $\dfrac{3}{4}$

Explanation :

METHOD 1:

In tossing three coins, the sample space is given by,

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

∴ n(S) = 8

Let E = event of getting at least 2 heads.

Then, E = {HHH, HHT, HTH, THH}

∴ n(E) = 4

∴ P(E) = $ \dfrac{\text{n(E)}}{\text{n(S)}}$

⇒ P(E) = $\dfrac{4}{8}$

⇒ P(E) = $\dfrac{1}{2}$

METHOD 2:

If n unbaised coins are tossed. Total number of outcomes in the sample space = 2ⁿ

∴ n(S) = 2³ = 8

Let E = event of getting at least 2 heads.

Then, E = {HHH, HHT, HTH, THH}

∴ n(E) = 4

∴ P(E) = $ \dfrac{\text{n(E)}}{\text{n(S)}}$

⇒ P(E) = $\dfrac{4}{8}$

⇒ P(E) = $\dfrac{1}{2}$ 

Explanation :

Here, n(S) = 6 x 6 = 36

Let E = event of getting two numbers whose product is an even.

Then, E = {(1,2),(1,4),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,,2),(3,4),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,2),(5,4),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

∴ n(E) = 27

∴ P(E) = $ \dfrac{\text{n(E)}}{\text{n(S)}}$

⇒ P(E) = $\dfrac{27}{36}$

⇒ P(E) = $\dfrac{3}{4}$