Probability: Two Independent Events

Suppose $A$ and $B$ are two independent events. Then $P(A$ $\textrm{and}$ $B)$ or more commonly $P(A \cap B) = P(A) \cdot P(B)$. In other words, to find the probability of two independent events occurring, you just multiply the two.

Alright that's great, but what makes two events "independent"? Two events are independent simply if the occurrence of one does not affect the occurrence of another.

Here, getting a head on the first toss does not affect the probability of getting a head on the second toss since this is a fair coin and the probability of getting a head for each toss is $\frac{1}{2}$. So the probability of getting two heads is $\frac{1}{2} \cdot \frac{1}{2}$.

Independent Events

If $2$ events ($A$ and $B$) are independent, they have NO effect on each other. This means that the outcome of event $A$ has absolutely NOTHING to do with event $B$.

Here's an example of $2$ independent events:

In this example, one event (rolling a $4$) has nothing to do with the other (flipping a coin and getting tails).

In other words, the events are independent. This means that the event of rolling a $4$ will remain completely unchanged regardless of whether or not we land on tails.

Let's try an example:

Two events are said to be independent if the probability of one event does not affect the probability of the other. For example, a coin flip and drawing a card from a standard deck of $52$ cards are independent events. Similarly, rolling a fair, six-sided die will have no influence on a subsequent roll of the same die, meaning that the rolls, or trials, are independent of one another. Suppose $A$ and $B$ are independent events. To find the probability of $A$ and $B$ both occurring, denoted $P(A \bigcap B)$, simply multiply their individual probabilities together. $P(A) \times P(B) = P(A \bigcap B)$

Although it may look more complex, finding probabilities of two events of different types follows the same concept.

The same concept applies to more than two independent events. For example, the probability of rolling a six-sided die three times and getting the same result on all three rolls is simply $(1/6) \times (1/6) \times (1/6) = 1/216$.

Lack of understanding of the independence of events has led to an idea known as the "gambler's fallacy." This idea states that if an event has happened with unusual frequency, it is more likely to not happen on the next trial. For example, it is extremely unlikely that flipping a fair coin $99$ times would yield $99$ consecutive tails. However, the chance of a heads on the $100$th flip remains $\frac{1}{2}$ and is not increased due to the frequency of the tails. This deceiving idea has led to many to incorrectly assume that a streak of bad luck will change course, leading to the name "gambler's fallacy." A particularly infamous example occurred at the Monte Carlo Casino in $1913$, when a roulette wheel landed in black (probability of $\frac{1}{2}$) $26$ times in a row, incorrectly leading many to bet the wheel was due to land in red on the next trial. Millions of francs were gambled away.

(Video) Probability of Independent Events

by MathHelp.com

MathHelp.com has a good video on how to find the probability of two independent events.

$\large{\text{Summary}}$

In the problem given, we want to find the probability of getting a heads and then another heads when we toss two coins. Since the coins are independent of each other, meaning the result of one coin will not affect the other, we can just multiply their individual probabilities together. $P(heads) = \frac{1}{2}$ for any standard, two-sided coin, so $P(heads, heads)$ will be:

$$P(heads, heads) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$$