A coin is tossed and a die thrown.
A coin having two sides have a total outcome of 2 viz. {H,T}
A die has 6 faces and will have a total outcome of 6 i.e. {1, 2,3,4,5,6}
Let P(A) be the probability of getting the number 6.
\(\therefore P(A)=\frac{1}{6}\)
Let P(B) be the probability of getting a head.
The sample space of B ={H}
\(\therefore P(B)=\frac{1}{2}\)
Let P(A ∩ B) be the probability of getting the number 6 and a head.
Tip – By conditional probability , P(A/B) = \(\frac{P(A \cap B)}{P(B)}\) where P(A/B) is the probability of occurrence of the event
A given that B has already occurred.
The probability that 6 came up given that head came up:
