Question

An experiment succeeds twice as often as it fails. Find the probability that in the next six trials there are at most 2 successes.
A. $ \dfrac{73}{729} $
B. $ \dfrac{72}{729} $
C. $ \dfrac{71}{729} $
D. $ \dfrac{70}{729} $

Answer 1

Hint: We first find the options which describe the event of getting at most 2 success out of 6 trials. From the relation of experiment succeeding twice as often as it fails, we find the individual probability. From the options of getting at most 2 success out of 6 trials, we find the total probability.

Complete step by step answer:
An experiment succeeds twice as often as it fails. This means out of x number of trials if the experiment fails y times then it gets success 2y times where $ y+2y=x\Rightarrow x=3y $ .
From this we find the probability of the experiment succeeding where we denote that event as event A. So, $ p\left( A \right)=\dfrac{2y}{3y}=\dfrac{2}{3} $ . Again, the probability of the experiment failing is $ p{{\left( A \right)}^{c}}=1-p\left( A \right)=1-\dfrac{2}{3}=\dfrac{1}{3} $ .
We need to find the probability that in the next six trials there are at most 2 successes.
At most 2 success in 6 trials means three options. They are getting 0 success out of 6 trials, 1 success out of 6 trials, 2 success out of 6 trials.
We individually find their probability.
The events of failing and succeeding are independent events.
For 0 success out of 6 trials, we have all 6 trials being failed.
The probability is $ {{\left( \dfrac{1}{3} \right)}^{6}}=\dfrac{1}{729} $ .
For 1 success out of 6 trials, we have all 5 trials being failed and 1 being succeeded. We need to find the trial number which gets successful.
The probability is $ {}^{6}{{C}_{1}}\left( \dfrac{2}{3} \right){{\left( \dfrac{1}{3} \right)}^{5}}=6\times \dfrac{2}{729}=\dfrac{12}{729} $ .
For 2 successes out of 6 trials, we have all 4 trials being failed and 2 being succeeded. We need to find the trial numbers which get success.
The probability is $ {}^{6}{{C}_{2}}{{\left( \dfrac{2}{3} \right)}^{2}}{{\left( \dfrac{1}{3} \right)}^{4}}=\dfrac{6!}{2!\times 4!}\times \dfrac{4}{729}=\dfrac{60}{729} $ .
The different options are exclusive events. So, we add them to get the total probability.
 $ \dfrac{1}{729}+\dfrac{12}{729}+\dfrac{60}{729}=\dfrac{73}{729} $ . The correct option is A.

Note:
The complementary event of the choices is getting 3, 4, 5, 6 successes out of 6 trials. Then we need to subtract that probability from 1 to find the correct answer of the problem. We don’t use that process to avoid complicated operations.