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Question Number 123389 by aurpeyz last updated on 25/Nov/20

a n u n b i a s e d c o i n i s t o s s e d t h r e e t i m e s i f a d i s c r e r e

r a n d o m v a r i a b l e X d e n o t e d t h e

n u m b e r o f t i m e s a t a i l a p p e a r s

1 . l i s t t h e s a m p l e s p a c e

2 . f i n d t h e d i s t r i b u t i o n o f X

3 . t h e m e a n o f X

Answered by som(math1967) last updated on 25/Nov/20

1 . (TTT, TTH, THT, HTT, THH

HHT, HTH, HHH)

probability of tail = p = \frac{1}{2}

probability of head = q = \frac{1}{2}

P (X) =^{n} c_{r} p^{r} q^{n - r}

i {No tail .

P (X = 0) =^{3} c_{0} {(\frac{1}{2})}^{0} {(\frac{1}{2})}^{3 - 0} = \frac{1}{8}

ii) one tail

P (X = 1) =^{3} c_{1} {(\frac{1}{2})}^{1} {(\frac{1}{2})}^{3 - 1} = \frac{3}{8}

iii) 2 tail = \frac{3}{8}

iv) 3 tail = \frac{1}{8}

2 . {\begin{cases} x & 0 & 1 & 2 & 3 \\ P (x) & \frac{1}{8} & \frac{3}{8} & \frac{3}{8} & \frac{1}{8} \end{cases}

3 . Mean \bar{x} = 0 \times \frac{1}{8} + 1 \times \frac{3}{8} + 2 \times \frac{3}{8}

3 \times \frac{1}{8} = \frac{12}{8} = 1.5