Help requested, conditional probability question, What is my mistake?

Question

Can anyone help me with this question

If the real answer is "no", then give 50% yes and 50% no If the real answer is "yes“, then give 60% yes and 40% no

Suppose that the noisy answers consist of 55% yes and 45% no Estimate the percentage of "yes" in the original answer

This was my attempt
$p(\text{yes | the real answer is "no"}) = 0.5$
$p(\text{no | the real answer is "no"}) = 0.5$
$p(\text{yes| the real answer is "yes"}) = 0.6$
$p(\text{no | the real answer is "yes"}) = 0.4$

$p(\text{yes}) = 0.55$
$p(\text{no}) = 0.45$

p(the real answer is "yes | yes) = p(yes| the real answer is "yes") / [p(yes| the real answer is "no") + p(yes| the real answer is "yes")] $= 0.6 / [0.5 + 0.6] = 0.54$

p(the real answer is "yes ^ yes) = p(the real answer is "yes | yes) x p(yes) $= 0.54 \times 0.55 = 0.3$

I feel what I am doing is wrong.

is $p(A|B) = p(B|A)/[p(B|A) + p(B|\text{not }A)]$ ?

Answer 1

Let $X$ be the event the real answer is "yes" and $Y$ be the event the noisy answer is "yes". The question tells you:

• $\mathbb P(Y \mid X^c)=0.5$ and $\mathbb P(Y^c \mid X^c)=0.5$
• $\mathbb P(Y \mid X)=0.6$ and $\mathbb P(Y^c \mid X)=0.4$
• $\mathbb P(Y)=0.55$ and $\mathbb P(Y^c )=0.45$

Since $$\mathbb P(Y) = \mathbb P(Y \mid X)\mathbb P(X)+ \mathbb P(Y \mid X^c)\mathbb P(X^c)$$ i.e. $$\mathbb P(Y) = \mathbb P(Y \mid X)\mathbb P(X)+ \mathbb P(Y \mid X^c)(1-\mathbb P(X))$$ you can say

$$\mathbb P(X)=\frac{\mathbb P(Y)- \mathbb P(Y \mid X^c)}{\mathbb P(Y \mid X)-\mathbb P(Y \mid X^c)}$$

which in this case is $\dfrac{0.55-0.5}{0.6-0.5}=0.5$

Answer 2

No, the formula at the end is not correct, and should be $$P(A\mid B)=\frac{P(B\mid A)P(A)}{P(B)}.$$ However, this formula isn't immediately useful here, as if $A$ is "real answer yes" and $B$ is "yes" then both $P(A\mid B)$ and $P(A)$ are unknown. Instead, try using $$P(B)=P(A)P(B\mid A)+(1-P(A))P(B\mid \neg A).$$