What is the expectation of number of failures preceding the first success in an infinite series of independent trials with constant probabilty 2/3?
Answers
Total expected number of failures = 1/2
Step-by-step explanation:
Probability of Success = 2/3
Probability of Failure = 1/3
Expected number of failure Probability
0 2/3
1 (1/3)(2/3)
2 (1/3)²(2/3)
3 (1/3)³(2/3)
n (1/3)ⁿ(2/3)
n tends to infinity
Total expected number of failures
= 0 * 2/3 + 1 *(1/3)(2/3) + 2* (1/3)²(2/3) + 3*(1/3)³(2/3)+...............+ n(1/3)ⁿ(2/3)
= (2/3) ( 1 * 1/3 + 2(1/3)² + 3(1/3)³ +................... n(1/3)ⁿ)
= (2/3)S
S = 1 * 1/3 + 2(1/3)² + 3(1/3)³ +................... n(1/3)ⁿ +.......
(1/3)S = (1/3)² + 2(1/3)³ +......................+(n-1)(1/3)ⁿ +....
=> (2/3)S = 1/3 + (1/3)² + (1/3)³ + ..............................+(1/3)ⁿ
=> (2/3)S = (1/3) /( 1- 1/3) ( sum of infinite GP a/(1 - r)
=> (2/3)S = (1/3) /(2/3)
=> (2/3)S = 1/2
=> S = 3/4
Total expected number of failures = (2/3)(3/4)
= 1/2
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