Statistics And Probability Question 20
Question: In a sequence of independent trials, the probability of success on each trial is ¼. The probability that the second success occurs on the fourth or later trial, if the trials continue up to the second success only, is
Options:
A) $ \frac{5}{32} $
B) $ \frac{27}{32} $
C) $ \frac{23}{32} $
D) $ \frac{9}{32} $
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] Let the second success occur at the nth trial. This means that there was exactly one success in the first n - 1 trials, so that the probability of getting the second success at the nth trial is $ p_{n}={{(}^{n-1}}C_1p{q^{n-1-1}})p=(n-1)p^{2}{q^{n-2}} $ Therefore the probability of the required event is $ p_4+p_5+p_6+…=3p^{2}q^{2}+4p^{2}q^{3}+5p^{2}q^{4}+6p^{2}q^{5}+… $ $ =p^{2}q^{2}(3+4q+5q^{2}+6q^{3}+…) $ $ =p^{2}q^{2}[3(1+q+q^{2}+q^{3}+…)+q(1+2q+3q^{2}+…)] $ $ =p^{2}q^{2}[(3{{(1-q)}^{-1}}+q{{(1-q)}^{-2}}]=p^{2}q^{2}( \frac{3}{p}+\frac{q}{p^{2}} ) $ $ =q^{2}(3p+q)=q^{2}(2p+p+q)=q^{2}(2p+1) $ $ ={{( \frac{3}{4} )}^{2}}[ 2( \frac{1}{4} )+1 ]=\frac{9}{12}( \frac{1}{2}+1 )=\frac{27}{32} $
BETA
