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 $k^{t h}$ roll of the die?

(ii) How many elements of the sample space correspond to the event that the 2 appears not later than the $k^{t h}$ roll of the die?

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Payal Gupta | Contributor-Level 10

8 months ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} N u m b e r o f S a m p l e s p a c e = 6 \\ (i) G i v e n t h a t 2 a p p e a r s o n t h e k t h r o l l o f t h e d i e . \\ S o f i r s t (k - 1) t h r o l l h a v e 5 o u t c o m e s e a c h a n d k t h r o l l r e s u l t s 2 i . e . o n l y 1 o u t c o m e . \\ ∴ N u m b e r o f e l e m e n t o f s a m p l e s p a c e c o r r e s p o n d t o t h e e v e n t t h a t 2 a p p e a r s o n t h e k t h \\ r o l l o f t h e d i e = 5^{k - 1} . \\ (i i) I n t h i s c a s e, 2 a p p e a r s n o t l a t e r t h a n k t h r o l l o f t h e d i e, t h e n i t i s p o s s i b l e t h a t 2 c o m e s \\ i n f i r s t t h r o w t h e n o u t c o m e s w i l l b e 5 a n d 2 o u t c o m e s i n second t h r o w i . e . 1 o u t c o m e . \\ ∴ P o s s i b l e o u t c o m e = 5 * 1 = 5 \\ S i m i l a r l y, i f 2 d o e s n o t a p p e a r i n second t h r o w a n d a p p e a r s i n t h i r d t h r o w \\ ∴ P o s s i b l e o u t c o m e = 5 * 5 * 1 \\ N o w w e h a v e t h e s e r i e s : \\ = 1 + 5 + 5 * 5 + 5 * 5 * 5 + \dots + 5^{k - 1} \\ = 1 + 5 + 5^{2} + 5^{3} + \dots + 5^{k - 1} \\ = \frac{1 . (r^{k} - 1)}{r - 1} = \frac{5^{k - 1}}{5 - 1} \\ = \frac{5^{k - 1}}{4} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y = \frac{5^{k - 1}}{4} . \end{array}$

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