An experiment consists of rolling a die until a 2 appears.

Question:

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?

Solution:

Given number of outcomes when die is thrown = 6

(i) Given that 2 appears on the kth roll of the die.

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

∴ Number of outcomes = 5k-1

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

= 5k-1

Possible outcomes if 2 appears before kth roll

= 1 + 5 + 52 + 53+ … + 5k-1

Here, we get the series

We know that,

$S_{n}=\left\{\begin{array}{l}\frac{a\left(r^{n}-1\right)}{r-1} \text { if } r>1 \\ \frac{a\left(1-r^{n}\right)}{1-r} \text { if } r<1\end{array}\right.$

So, here $a=1$

and $r=\frac{5}{1}=5>1$

Hence,

$=\frac{1 \times\left(5^{k}-1\right)}{5-1}$

$=\frac{5^{k}-1}{4}$

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now