In this lecture we’ll look at the conjunction rule, the rule for calculating the probability of a conjunction of events. But we’ll deal with a special case, where the events in question are independent, and the rule takes on a very simple form.
Let’s consider coin tosses again. The probability of a single coin landing heads is 1/2, right?
Now what if we toss two coins at the same time? What is the probability that both coins will land heads?
Most people will see the answer right away, because we’re familiar with these sorts of cases. We know that the probability of both landing heads is going to be less than the probability of just one landing heads. When we’re dealing fractions, we know that multiplying fractions gives us a smaller number.
In this case, if we multiply the probabilities of the two independent events, we get the right answer: 1/2 times 1/2 equal to 1/4. There’s a 25% chance of both coins landing heads.
If we go for three coins, it’s the same idea. We just multiply the probabilities of each individual event, and the answer is half as small again, 1/8.
This is the restricted conjunction rule:
P(A and B) = P(A) x P(B)
If A and B are independent events, the probability of the conjunction of two events, which is just the probability of the two events both occurring, or of the corresponding propositions both being true, is just the product of the probabilities taken separately.
If the events are not independent, then we need to use a modified version of this rule, but we’ll take that up in the next lecture.
Let’s just look at one more example.
Using a six-sided dice, what is the probability of rolling three sixes in a row? That is, what is
P(6 and 6 and 6) = ?
Dice rolls, like coin tosses, are independent events, so we can use the restricted conjunction rule. The probability of rolling a six is just one in six. So the probability of rolling three sixes in a row is just 1/6 x 1/6 x 1/6, which is one in 216:
P(6 and 6 and 6)
= P(6) x P(6) x P(6)
= (1/6) x (1/6) x (1/6)
This is a pretty simple rule, but before we leave I want you to think about what this rule means from the sample space perspective. In this framework, events are associated with subsets of the sample space, and probabilities are associated with the ratios of the corresponding subsets to the total space of possible outcomes.
Here’s the sample space for single coin toss:
It’s just a listing of the set of possible elementary outcomes, and in this case there are just two, heads and tails. The probability of the coin landing heads is equal to the ratio of outcomes where it lands heads, divided by the total number of elements in the sample space, which in this case is obviously just 1 out of 2 or 1/2.
Here’s the sample space for the event where two coins are tossed:
You’ve got twice as many possible combinations of outcomes — heads heads, heads tails, tails heads, and tails tails. This sample space has four elementary outcomes, and the probability of two heads is just the ratio of the number of elements where the outcome is two heads, divided by the total number of elementary outcomes, which is 4. So the probability for this event is just 1 in 4.
Here’s the sample space for three coin tosses:
There are eight possible combinations of heads and tails, and the probability of landing three heads is just 1/8.
Here’s the sample space for rolling two dice:
There are 36 possible outcomes. The probability of rolling two sixes is 1/6 times 1/6, or 1 in 36.
It’s rare that you’ll have to write out the sample space like this to solve a problem, but it’s a helpful reminder of what the mathematical rule means, and why it gives us the right answers.