Given P(A) = \(1\over2\), P(B) = \(1\over3\) and \(P(A\cap B)={1\over6}\)  Are the events A and B independent?

1

(a)

\(P(A) \cdot P(B)=\frac{1}{2} \cdot \frac{1}{3}=\frac{1}{6}=P(A \cap B)\)
Yes,the events are independent.

Bayes’ Theorem If E₁ , E₂ ,..., E_n are n non empty events which constitute a partition of sample space S, i.e. E₁ , E₂ ,..., E_n are pairwise disjoint and E₁∪ E₂∪ ... ∪ E_n = S and A is any event of nonzero probability, then
\(\mathrm{P}\left(\mathrm{E}_i \mid \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_i\right) \mathrm{P}\left(\mathrm{A}_{\mid} \mathrm{E}_i\right)}{\sum_{j=1}^n \mathrm{P}\left(\mathrm{E}_j\right) \mathrm{P}\left({\left.\mathrm{A} \mid E_j\right)}_1\right.} \text { for any } i=1,2,3, \ldots, n\)

1

(a)

p = 0.5

Given P(A) = 0.2, P(B) = 0.3 and \(P(A\cap B)=0.3\) Find P(A/B)

1

(a)

1/3

Given P(A) = 0.4, P(B) = 0.7 and P(B/A) = 0.6, Find \(P(A\cup B)\)

1

(a)

\(P(B / A) =\frac{P(A \cap B)}{P(A)} \)
\(\Rightarrow 0.6 \times 0.4 =P(A \cap B)\)
\(\Rightarrow P(A \cap B) =0.24 \)
\(\text { Now, } \ P(A \cup B) =P(A)+P(B)-P(A \cap B) \)
\(=0.4+0.7-0.24=0.86 \)

Events E and F are given to be independent. Find P(F) if it is given that P(E) = 0.60 and P(E\(\cap\)F) = 0.35

1

(a)

For independent events,
\( P(E \cap F) =P(E) \cdot P(F) \)
\(\Rightarrow 0.35 =0.60 \times P(F) \Rightarrow P(F)=\frac{7}{12}=0.58 \)

Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.

1

(a)

No