An experiment consists of rolling a die until a 2 appears.
(i) How many elements of the sample space correspond to the event that the 2 appears on the kth roll of the die?

(ii) How many elements of the sample space correspond to the event that the 2 appears not later than the kth roll of the die?

Question

Philoid · Accepted Answer

Number of outcomes when die is thrown = 6

(i) Given that 2 appears on the k^th roll of the die.

So, first (k – 1)^th roll have 5 outcomes each and k^th roll results 2

∴ Number of outcomes = 5^k-1

(ii) If we consider that 2 appears not later than k^th roll of the die, then 2 comes before k^th roll.

If 2 appears in first roll, number of ways = 1 outcome

If 2 appears in second roll, number of ways = 5 x 1 (as first roll does not result in 2)

If 2 appears in third roll, number of ways = 5 x 5 x 1 (as first two rolls do not result in 2)

Similarly, if 2 appears in (k – 1)^th roll, number of ways

= (5 x 5 x 5 … (k- 1) times) x 1

= 5^k-1

Possible outcomes if 2 appears before k^th roll

= 1 + 5 + 5² + 5³+ … + 5^k-1

Here, we get the series

We know that,

So, here a = 1

and

Hence,

An experiment consists of rolling a die until a 2 appears.(i) How many elements of the sample space correspond to the event that the 2 appears on the kth roll of the die?(ii) How many elements of the sample space correspond to the event that the 2 appears not later than the kth roll of the die?