7.1: Independent Events
- Page ID
- 139286
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Suppose we flipped a coin and rolled a die, and wanted to know the probability of getting
a head on the coin and a 6 on the die.
Solution
We could list all possible outcomes: \(\{\mathrm{H} 1, \mathrm{H} 2, \mathrm{H} 3, \mathrm{H} 4, \mathrm{H} 5, \mathrm{H} 6, \mathrm{~T} 1, \mathrm{~T} 2, \mathrm{~T} 3, \mathrm{~T} 4, \mathrm{~T} 5, \mathrm{~T} 6\}\) Notice there are \(2 \cdot 6=12\) total outcomes. Out of these, only 1 is the desired outcome, so the probability is \(\mathrm{P}(\) head \()=\dfrac{1}{12}\).
Events A and B are independent events if the probability of Event B occurring is the same whether or not Event A occurs.
Are these events independent?
a) A fair coin is tossed two times. The two events are (1) first toss is a head and (2) second toss is a head.
b) You draw a card from a deck, then draw a second card without replacing the first.
Solution
- The probability that a head comes up on the second toss is \(1 / 2\) regardless of whether or not a head came up on the first toss, so these events are independent.
- The probability of the first card being red is \(P\) (first card is red) \(=\dfrac{26}{52}=\dfrac{1}{2}\). Since the first card is not returned to the deck, the probability of the second card being red is not the same as the probability that the first card is red. After drawing the first card, there are only 51 cards remaining in the deck. If the first card was red, then there are only 25 red cards remaining, and the probability that the second card would be red is \(\mathrm{P}(\) second card is red \()=\dfrac{25}{51}\). If the first card was not red, then the probability that the second card would be red is \(\mathrm{P}(\) second card is red \()=\dfrac{26}{51}\). These events are NOT independent.
When two events are independent, the probability of both occurring is just the product of the probabilities of the individual events.
If events \(A\) and \(B\) are independent, then the probability of both \(A\) and \(B\) occurring is
\[
P(A \text { and } B)=P(A) \cdot P(B)
\]
where \(P(A\) and \(B)\) is the probability of events \(A\) and \(B\) both occurring, \(P(A)\) is the probability of event \(A\) occurring, and \(P(B)\) is the probability of event \(B\) occurring
If you look back at the coin and die example from earlier, you can see how the number of outcomes of the first event multiplied by the number of outcomes in the second event multiplied to equal the total number of possible outcomes in the combined event.
In Alex’s drawer she has 10 pairs of socks, 6 of which are white, and 7 tee shirts, 3 of which are white. If she randomly reaches in and pulls out a pair of socks and a tee shirt, what is the probability that both are white?
Solution
The probability of choosing a white pair of socks is \(\dfrac{6}{10}\).
The probability of choosing a white tee shirt is \(\dfrac{3}{7}\).
The probability of both being white is \(\dfrac{6}{10} \cdot \dfrac{3}{7}=\dfrac{18}{70}=\dfrac{9}{35}\)
A couple has two children. Find the probability that both children are girls.
- Answer
-
1/4
You roll two die. Find the probability that the first die lands on an even number, and the second die lands on 5.
Solution
Since the outcome of the first die has no effect on the outcome of the second die, the events are independent.
\(\mathrm{P}(\) first die is even \()=\dfrac{3}{6}=\dfrac{1}{2}\)
\(\mathrm{P}(\) second die lands on 5\()=\dfrac{1}{6}\)
\(\mathrm{P}(\) first die is even AND second die lands on 5\()=\dfrac{1}{2} \times \dfrac{1}{6}=\dfrac{1}{12}\)
As a check, it may be helpful to consider all possible outcomes in the experiment of rolling a pair of dice:
|
1,1 |
1,2 |
1,3 |
1,4 |
1,5 |
1,6 |
|
2,1 |
2,2 |
2,3 |
2,4 |
2,5 |
2,6 |
|
3,1 |
3,2 |
3,3 |
3,4 |
3,5 |
3,6 |
|
4,1 |
4,2 |
4,3 |
4,4 |
4,5 |
4,6 |
|
5,1 |
5,2 |
5,3 |
5,4 |
5,5 |
5,6 |
|
6,1 |
6,2 |
6,3 |
6,4 |
6,5 |
6,6 |
Notice there are 3 outcomes (out of 36 possible outcomes) where the first die is even and the second die lands on 5 . We could have found this probability directly as \(\dfrac{3}{36}=\dfrac{1}{12}\).