Question

Question:

(a) When are two events \(E\) and \(F\) independent?

(b) Suppose \(E\) is the event that an even number appears when a fair die is rolled, and \(F\) is the event that a 5 or 6 comes up. Are \(E\) and \(F\) independent?

Short answer

Answer

(a) The resultant answer is two events are independent, if the probability that one event occurs in no way affects the probability of the other event occurring.

(b) The resultant answer \(E\) and \(F\) are independent.

Step-by-step solution

Given data

The given data is event \(E\) and \(F\).

Find the probabilityof the event

(a)

Two events are independent, if the probability that one event occurs in no way affects the probability of the other event occurring; \(P(A \cap B) = P(A)P(B)\).

Find the probability of the event 

(b)

\(E = \)even number

\(F = 5{\kern 1pt} ,6\)

When rolling a single die, then there are 6 possible outcomes: 1,2,3,4,5,6.

3 of the 6 possible outcomes are even \((2,4,6)\).

The probability is the number of favorable outcomes divided by the number of possible outcomes:

\(p(E) = \frac{{\# {\rm{ of favorable outcomes }}}}{{\# {\rm{ of possible outcomes }}}} = \frac{3}{6}\)

2 of the 6 possible outcomes are 5 or 6.

\(p(F) 7= \frac{{\# {\rm{ of favorable outcomes }}}}{{\# {\rm{ of possible outcomes }}}} = \frac{2}{6}\)

1 of the 6 possible outcomes are even and 5 or 6 (that is, 6).

\(\begin{aligned}{}p(E \cap F) = \frac{{\# {\rm{ of favorable outcomes }}}}{{\# {\rm{ of possible outcomes }}}}\\ = \frac{1}{6}\end{aligned}\)

We now note;

\(\begin{aligned}{}p(E \cap F) &= \frac{1}{6}\\ &= \frac{2}{6} \cdot \frac{3}{6}\\ &= p(E)p(F)\end{aligned}\)

which implies that \(E\) and \(F\) are independent.