Probability that A speaks truth is \(\frac{4}{5}\).A coin is tossed.A report that a head appears.The probability that actually there was head is:
(\(\frac{4}{5}\))
(\(\frac{1}{2}\))
(\(\frac{1}{5}\))
(\(\frac{2}{5}\))
Let A be the event that the man reports that head occurs in tossing a coin and let E1 be the event that head occurs and E2 be the event head does not occur.
P(E1)=\(\frac{1}{2}\),P(E2)=\(\frac{1}{2}\)
P(A|E1)=P(A reports that head occurs when head had actually occur red on the coin)=\(\frac{4}{5}\)
P(A|E2)=P(A reports that leads occurs when head had not occur red on the coin)=1-\(\frac{4}{5}\)=\(\frac{1}{5}\)
By Bayes'theorem,
P(E1|A)=\(\frac{P(E1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}\)=\(\frac{\frac{1}{2}×\frac{4}{5}}{\frac{1}{2}×\frac{4}{5}+\frac{1}{2}×\frac{1}{5}}\)=4/4+1=4/5
Hence,option (A) is correct.
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, find:
\(Evaluate \ P(A∩B)\ if \ 2P(A) = P(B) =\) \(\frac {5}{13}\) \(and \ P(A|B)=\) \(\frac 25\)
Bayes’ Theorem is a part of the conditional probability that helps in finding the probability of an event, based on previous knowledge of conditions that might be related to that event.
Mathematically, Bayes’ Theorem is stated as:-
\(P(A|B)=\frac{P(B|A)P(A)}{P(B)}\)
where,
This formula confines well as long as there are only two events. However, Bayes’ Theorem is not confined to two events. Hence, for more events.