Question

A person throws a dice while he gets a number greater than 2. The probability that he gets a 6 in the last thrown is

• A. 2/3
• B. ¼
• C. 1/3
• D. 1/12

Answer 1

Answer: (D)

1/12

Answer 2

Let E₁ = the even that six shows when a dice is thrown

E₂ = the number less than or equal to 2 shows where a dice is thrown

P(E₁) = 1/6 and P(E₂) = 2/6 = 1/3

E = the event that six turns up in the last thrown = the event that E₁ happen in the all previous throws and E₂ happens in the last throw P(5) = P(E₁E₂ or E₁E₁E₂ or E₁E₁E₁E₂.....) = P(E₁). P(E₂) + P(E₁). P(E₁). P(E₂) + P(E₁). P(E₁). P(E₁). P(E₂) +......= $\frac { 1 } { 3 } \cdot \frac { 1 } { 6 } + \left( \frac { 1 } { 3 } \right) ^ { 2 } \cdot \frac { 1 } { 6 } + \left( \frac { 1 } { 3 } \right) ^ { 3 } \cdot \frac { 1 } { 6 }$ +........ P(5) = $\frac { 1 } { 3 } \cdot \frac { 1 } { 6 } \times \frac { 1 } { 1 - 1 / 3 }$ Ž P(5) = $\frac { 1 } { 12 }$