Question

A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. The probability that it is actually a six is

Answer 1

Answer:

mark me as brainliest and support me to give more valuable answers.

Step-by-step explanation:

Answer 2

Answer:

$\boxed{\sf \: Probability\:that\:it\:actually\:a\:6\:is\:\dfrac{3}{8} \: }$

Step-by-step explanation:

Let assume the following events

E₁ : die show a number 6.

E₂ : die do not show a number 6

E : a man reports that 6 occurs on the die.

Now,

$\sf \: P(E_1) = \dfrac{1}{6}$

$\sf \: P(E_2) = \dfrac{5}{6}$

Now,

$\sf \: P(E \: \mid \: E_1) = \dfrac{3}{4}$

and

$\sf \: P(E \: \mid \: E_2) = \dfrac{1}{4}$

Now, By definition of Bayes Theorem, we have

$\sf P(E_1 \mid \: E) = \dfrac{P(E_1) . P(E \mid E_1)}{P(E_1) . P(E \mid E_1) + P(E_2) . P(E \mid E_2)}$

So, on substituting the values, we get

$\sf P(E_1 \mid \: E) = \dfrac{\dfrac{1}{6} \times \dfrac{3}{4} }{\dfrac{1}{6} \times \dfrac{3}{4} + \dfrac{5}{6} \times \dfrac{1}{4} }$

$\sf P(E_1 \mid \: E) = \dfrac{3}{3 + 5}$

$\sf\implies \sf P(E_1 \mid \: E) = \dfrac{3}{8}$

Hence,

$\sf\implies \sf Probability\:that\:it\:actually\:a\:6\:is\:\dfrac{3}{8}$