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  • 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

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?

Answers (1)

We know that, the no. of outcomes when a die is thrown is 6

  1. 2 appear on the kth roll of the die     ……… (given)

Thus, the first (k – 1)th roll has 5 outcomes each

Thus, no. of outcomes = 5k-1

  1. Let us consider that 2 come before the kth roll of the die and not after that.

Thus,

In the first roll, no. of ways in which 2 appears will be = 1 outcome

In the second roll, no. of ways in which 2 appears will be = 5 x 1 outcome

……. (since the first roll doesn’t result in 2)

In the third roll, no. of ways in which 2 appears will be = 5 x 5 x 1 outcome

……. (since the first two rolls doesn’t result in 2)

In the (k – 1)th roll, no. of ways in which 2 appear will be = [5 x 5 x 1 ….. (k – 1)] outcome

                                                                                    = 5k-1

Now, the possibility of 2 appearing before kth roll = 1 + 5 + 52 + 53 + …… + 5k-1

Now,

s_{n}=\left\{\begin{matrix} \frac{a\left ( r^{n}-1 \right )}{r-1},\, if\, \, r> 1 \\ \frac{a\left ( 1-r^{n} \right )}{1-r},\, if\, \, r< 1 \end{matrix}\right.

Thus, here, a = 1 & r = 5/1 = 5 >1

Thus,

= 1 x (5k – 1) / 5 – 1

= 5k – 1 / 4

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