AP EAPCET 21 May 2025 Shift 1 Paper

The person tells the truth 3 out of every 4 times.
Let's define:
- A : The die really shows six.
- B : The die does not show six.
- E : The person says the die shows six.
The chance of rolling a six when throwing a die, P(A), is ‌

1

6

.
The chance of not rolling a six, P(B), is ‌

5

6

.
If a six really comes up, the chance the person will say "six" (tells the truth) is P(E∣A)=‌

3

4

.
If a six did NOT come up, the chance the person wrongly says "six" (tells a lie) is P(E∣B)=‌

1

4

.
We want to find: What is the chance the die REALLY shows six if the person says it is a six? This is P(A∣E).
Use Bayes' Theorem:
P(A∣E)=‌

P(A)⋅P(E∣A)

P(A)⋅P(E∣A)+P(B)⋅P(E∣B)

Put in the values:
P(A∣E)=‌

‌ 1 6 × 3 4

1 6 ×‌ 3 4 +‌ 5 6 ×‌ 1 4

Solve:
P(A∣E)=‌

‌ 3 24

3 24 +‌ 5 24

=‌

3

8