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Home » 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 roll of the die? (ii) How many elements of the sample space correspond to the event that the 2 appears not later than the roll of the die?
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 k^{th} roll of the die?
(ii) How many elements of the sample space correspond to the event that the 2 appears not later than the k^{th} roll of the die?
CBSE | Class 11 | Maths | NCERT Exemplar | Probability
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 k^{th} roll of the die?
(ii) How many elements of the sample space correspond to the event that the 2 appears not later than the k^{th} roll of the die?

Solution:
The given no. of outcomes when die is thrown =6
(i) If 2 appears on the k^{th} roll of the die.
Therefore, the first (k-1) roll have 5 outcomes each and K^{th} roll results 2
No. of outcomes =5^{k-1}
(ii) If it is considered 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, no. of ways =1
If 2 appears in second roll, no. of ways
=5 \times 1 (as first roll doesn’t result in 2)
If 2 appears in third roll, no. of ways
=5 \times 5 \times 1 (as first two rolls doesn’t result in 2 )
In the similar way if 2 appears in (k-1)^{th} roll, no. of ways =[5 \times 5 \times 5 \ldots(k-1) times ] \times 1=5^{k-1}
The possible outcomes if 2 appears before k^{th} roll are =1+5+5^{2}+5^{3}+\ldots+5^{k-1}
=\frac{1\left(5^{k}-1\right)}{5-1}=\frac{5^{k}-1}{4}

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