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An experiment consists of rolling a die until a 2 appears. 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? - Mathematics

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प्रश्न

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

योग
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उत्तर

Number of sample space = 6

In this case, 2 appears not later than kth roll of the die

Then it is possible that 2 comes in first throw i.e. 1 outcome.

If 2 does not appear in first throw

Then outcomes will be 5 and 2 outcomes in second throw i.e. 1 outcome.

∴ Possible outcome = 5 × 1 = 5

Similarly, if 2 does not appear in second throw and appears in third throw

∴ Possible outcome = 5 × 5 × 1

Now we have the series:

= `1 + 5 + 5 xx 5 + 5 xx 5 xx 5 + ... + 5^(k - 1)`

= `1 + 5 + 5^2 + 5^3 + ... + 5^(k - 1)`

= `(.*(r^k - 1))/(r - 1)`

= `(5^k - 1)/(5 - 1)`

= `(5^k - 1)/4`

Hence, the required answer = `(5^k - 1)/4`.

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अध्याय 16: Probability - Exercise [पृष्ठ २९६]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
अध्याय 16 Probability
Exercise | Q 4.(ii) | पृष्ठ २९६
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