# Binomial distribution problem with conditioning on an event

A student is getting ready to take oral examination and is concerned about the possibility of having an ‘on’ day or an ‘off’ day. He figures that if he has an on day, then each of his examiners will pass him independently of each other with probability 0.8, whereas, if he has an off day, this probability will be reduced to 0.4. Suppose that the student will pass the examination if a majority of the examiners pass him. If the student feels that he is twice as likely to have an on day as an off day, should he request an examination with 3 examiners or with 5 examiners?

## Attempt

Let $A$ be the event that the student passes the oral examination. Let $B$ be the event that student has an 'on' day and then $B^c$ is the event that the student has an 'off' day. We know $P(B)=2P(B^c)$. Therefore,

$$P(B) = 2(1-P(B)) \implies 3 P(B) = 2 \implies P(B) = \frac{2}{3}$$

Now, by the law of total probability, one has

$$P(A) = P(A|B)P(B) + P(A|B^c)P(B^c) = \frac{2}{3} P(A|B) + \frac{1}{3} P(A|B^c)$$

Since each examiner either passes him or don't, we can call $X$ to be the number of examiners that passes him.

First case when the number of examiners is $3$, then we want to find, and also given that the students has taken a day 'on',

$$P(A|B)= P(X=2)+P(X=3) = {3 \choose 2 } 0.8^2 0.2 + {3 \choose 3} 0.8^3$$

and

$$P(A|B^c) = P(X=2)+P(X=3) = {3 \choose 2 } 0.4^2 0.6 + {3 \choose 3} 0.4^3$$

Now, replace everything above into the equations for $P(A)$.

Now, before I procced to do the case when we have $5$ examiners, is this the correct approach to this problem?

• I don't see any issue. – user625 Jun 18 '18 at 1:05
• In R: (2/3)*sum(dbinom(2:3, 3, .8)) + (1/3)*sum(dbinom(2:3, 3, .4)) returns 0.7146667, whereas (2/3)*sum(dbinom(3:5, 5, .8)) + (1/3)*sum(dbinom(3:5, 5, .4)) returns 0.7338667. – BruceET Jun 18 '18 at 1:23

Probability of off day $=\dfrac23$
Probability of on day $=\dfrac13$
If he request $3$ examinations then:
$P[$ pass if $2$ or more examiners pass hin$]$ $$=\dfrac23\left[\binom{3}{2}(0.4)^2\cdot(0.6)+\binom{3}{3}(0.4)^3\right]+\dfrac13\left[\binom{3}{2}(0.8)^2\cdot(0.2)+\binom{3}{3}(0.8)^3\right]$$ $$=\dfrac23\left[0.288+0.064\right]+\dfrac13[0.384+0.512]$$ $$=0.2347+0.2987$$ $$=0.533$$ Suppose, if he requests $5$ Examiners then:
$P[$pass if $3$ or more examiners pass him$]$
$P($pass$)=$ $$=\dfrac23\left[\binom{5}{3}(0.4)^3\cdot(0.6)^2+\binom{5}{4}(0.4)^4\cdot(0.6)+\binom{5}{5}(0.4)^5\right]+\dfrac13\left[\binom{5}{3}(0.8)^3\cdot(0.2)^2+\binom{5}{4}(0.8)^4\cdot(0.2)+\binom{5}{5}(0.8)^5\right]$$ $$=\dfrac23[0.2304+0.0768+0.01024]+\dfrac13[0.2048+0.4096+0.32768]$$ $$=0.2117+0.314$$ $$=0.526$$ Probability of passing is more, if there are $3$ examiners. So, he should prefer $3$ examiners.