Question

An unbiased die is tossed until a number greater than 4 appears. The probability that an even number of tosses needed is:

• A. ½
• B. 2/5
• C. 1/5
• D. 2/3

Answer 1

Answer: (B)

2/5

Answer 2

We have P(S) = P[5, 6] = $\frac { 2 } { 6 } = \frac { 1 } { 3 }$ .

Let S denote the occurrance of a number greater than 4 in a single throw of the die and F denote its failure.

⇒ P(F) = $\frac { 2 } { 3 }$ .

P (an even number of tosses is need).

= P(FS or FFFS or FFFFFS or ....)

= P(F) P(S) + P(F)³ P(S) + P(F)⁵ P(S) + ....

.