Independent Events Probability

When you toss a coin you get either heads or tails. Perhaps you've tossed the coin three times and it landed on heads those three times. You might think there's a chance for it to land on tails when you toss it the fourth time, but that's not true.

The fact that it has been landing on heads doesn't mean that you might get lucky and get a tail next time. Getting heads and getting a tail when a coin is tossed are two independent events.

Consider independent events A and B. P(A) is 0.7 and P(B) is 0.5, then:

\(P(A \cap B) = 0.7 \cdot 0.5 = 0.35\)

Consider two independent events A and B that involve rolling a die. Event A is rolling an even number and event B is rolling a multiple of 2. What is the probability of both events happening at the same time?

Solution

We have two events A and B.

Event A - rolling an even number

Event B - rolling a multiple of 2

Both events are independent. A die has six sides and the possible numbers to appear are 1, 2, 3, 4, 5, and 6. We are asked to find the probability of both events happening at the same time which is the intersection of both.

The formula to use is:

\(P (A \cap B) = P (A) \cdot P(B)\)

From the formula, we can see that to calculate the intersection, you need to know the probability of each event happening.

\[\text{Probability of an event happening} = \frac{\text{Number of ways the event can happen}}{\text{Number of possible outcomes}}\]

Therefore

\(P(A) = \frac{3}{6} = \frac{1}{2}\)

\(P(B) = \frac{3}{6} = \frac{1}{2}\)

We will now substitute the formula

\(P (A \cap B) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}\)

So the probability of both events happening is \(\frac{1}{4}\).

\(P(A) = 0.80\) and \(P(B) = 0.30\) and A and B are independent events. What is \(P(A \cap B)\)?

Solution

We are asked to find \(P(A \cap B)\) when \(P(A) = 0.80\) and \(P(B) = 0.30\). All we have to do is substitute into the formula below.

\(P (A \cap B) = P(A) \cdot P(B) = 0.80 \cdot 0.30\)

\(P(A \cap B) = P(A) \cdot P(B) = 0.80 \cdot 0.30\)

Therefore, \(P(A \cap B) = 0.24\)

In a classroom, 65% of the students like mathematics. If two students are chosen at random, what is the probability that both of them like mathematics and what is the probability that the first student likes mathematics and the second doesn't?

Solution

We have two questions here. The first is to find the probability of both students liking mathematics and the other is to find the probability of one liking mathematics and the other not liking it.

One student liking mathematics does not have affect on whether the second student likes mathematics too. So they are independent events. The probability of both of them liking mathematics is the probability of the intersection of the events.

If we call the events A and B, we can calculate using the formula below.

\(P(A \cap B) = P(A) \cdot P(B) = \frac{65}{100} \cdot \frac{65}{100}\)

Notice we divided by 100. This is because we are dealing with percentages.

Now, to find the probability of the first student liking mathematics and the second not liking it. These two are separate independent events and to find what we are looking for, we have to find the intersection of both events.

The probability of the first student liking mathematics is

\(P(A) = 65\% = 0.65\)

The probability of the second student not liking mathematics is

\(P(B) = 1- 0.65 = 0.35\)

We will now get our final answer by substituting the equation above.

\(P(A \cap B) = P(A) \cdot P(B) = 0.65 \cdot 0.35\)

C and D are events where \(P(C) = 0.50, \space P(D) = 0.90\). If \(P(C \cap D) = 0.60\), are C and D independent events?

Solution

We want to know if events C and D are independent. To know this, we will use the formula below.

\(P(C \cap D) = P(C) \cdot P(D)\)

We are given

If we substitute in the formula and we get the intersection to be something different from what the question suggests, then the events are not independent otherwise, they are independent.

Let's substitute.

\(P(C \cap D) = 0.50 \cdot 0.90 \quad P(C \cap D) = 0.45\)

We got 0.45 and the question says the intersection should be 0.60. This means the events are not independent.

A and B are independent events where \(P(A) = 0.2\) and \(P(B) = 0.5\). Draw a Venn diagram showing the probabilities for the event.

Solution

The Venn diagram needs some information to be put in it. Some of them have been given and we have to calculate for others.

\(P(A) = 0.2 \quad P(B) = 0.5 \quad P(A \cap B) = ? \quad P(S) = ? \space \text{(probability of the entire space)}\)

Now let's find the missing information.

\(P(A \cap B) = P(A) \cdot P(B) = 0.2 \cdot 0.5 = 0.1\)

\(P(S) = 1 - (P(A) + P(A \cap B) + P(B)) = 1-(0.2 + 0.1 +0.5) = 1-0.8 = 0.2\)

Now, let's draw the Venn diagram and put in the information.

From the Venn diagram below, find

1. \(P(C \cap D)\)
2. \(P(C \cup D)\)
3. \(P(C \cup D')\)

Solution

a. \(P(C \cap D)\)

\(P(C \cap D) = P(C) \cdot P(D)\)

From the Venn diagram,

So we will now substitute the formula.

\(P(C \cap D) = P(C) \cdot P(D) = 0.2 \cdot 0.6 = 0.12\)

b. \(P(C \cup D)\)

Here, we are to find the union of both events. This will be the summation of the probability of C, D and the intersect.

c. \(P(C \cup D')\)

So we have:

\(P(C \cup D') = P(C) + P(C \cap D) + S = 0.2 +0.12 + 0.08 = 0.4\)