Question

An ordinary dice is rolled a certain number of times. The probability of getting an odd number 2 times is equal to the probability of getting an even number 3 times. Then the probability of getting an odd number an odd number of times, is -

• A. $\frac { 1 } { 32 }$
• B. $\frac { 5 } { 16 }$
• C. $\frac { 1 } { 2 }$
• D. None of these

Answer 1

Answer: (C)

$\frac { 1 } { 2 }$

Answer 2

The probability of getting an odd number in a throw= $\frac { 3 } { 6 }$

= $\frac { 1 } { 2 }$ .

The probability of getting an odd number 2 times in n trails

= ⁿC₂ . $\left( \frac { 1 } { 2 } \right) ^ { 2 }$.

Similarly,the probability of getting an even number 3 times in n trails= ⁿC₃ . $\left( \frac { 1 } { 2 } \right) ^ { 3 }$. $\left( \frac { 1 } { 2 } \right) ^ { \mathrm { n } - 3 }$

From the question, ⁿC₂ . = ⁿC₃ . Ž n = 5

\ the required probability

= ⁵C₁ . $\left( \frac { 1 } { 2 } \right) ^ { 4 }$+ ⁵C₃ .$\left( \frac { 1 } { 2 } \right) ^ { 3 }$. $\left( \frac { 1 } { 2 } \right) ^ { 2 }$+ ⁵C₅ $\left( \frac { 1 } { 2 } \right) ^ { 5 }$

= (5 + 10 + 1). $\frac { 1 } { 2 ^ { 5 } }$= $\frac { 16 } { 32 }$ = $\frac { 1 } { 2 }$ Hence (3) is correct answer