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Independence and the Multiplication Rule

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Independence and the Multiplication Rule

Presentation 4.4

Overview and Examples

- Two events are independent if knowing that one occurs does NOT change the probability that the other occurs.
- Examples
- Successive tosses of a fair coin. Say you toss the coin four times and it comes up tails for the fourth straight time. What is the probability that it comes up heads on the fifth toss?
- Well, as the fair coin has no memory of what it has flipped in the past, it must still be 0.5. Each flip is independent and has NO bearing or influence on the next flip!

- Drawing two cards from a deck of cards. What is the probability of the second card being a heart?
- This depends on whether the first card was a heart or not.
- If the first card was a heart, there are now only 12 hearts left out of the 51 cards left.
- If the first card was not a heart, there are still 13 hearts left out of the 51 cards left.
- Therefore, these events (drawing the first card, then the second) are NOT independent!

- Successive tosses of a fair coin. Say you toss the coin four times and it comes up tails for the fourth straight time. What is the probability that it comes up heads on the fifth toss?

- Examples

- If events A and B are independent, then:
- P(A and B) = P(A)P(B)
- That is, if you have independence, you can simply multiply the probabilities of A and B to get the probability of A and B.

- The first Addition Rule ONLY applied to disjoint events.
- The General Addition rule applies to all events:
- P(A or B) = P(A) + P(B) – P(A and B)

- Consider the events from choosing a card from a deck of cards.
- Event A is choosing a jack
- Event B is choosing a spade
- What is the probability of choosing a jack or a spade?

Event A: Choosing a jack

P(A) = 4/52

Event B: Choosing a spade

P(B) = 13/52

If we simply added P(A) + P(B) to get 4/52 + 13/52 we would count the jack of spades TWICE!

These two events are NOT disjoint, since the jack of spades is an outcome in common!

Therefore use the general addition rule!

P(A or B) = P(A) + P(B) – P(A and B)

= 4/52 + 13/52 – 1/52

= 16/52 or .3077

Since the probability of getting a jack AND a spade is 1 out of 52 (since only one exists in the entire deck).

- A string of Holiday Lights for a tree contains 25 bulbs. Each of the bulbs function independently. If each bulb has a 0.015 chance of failing over a 3-year period, what is the probability that the entire string will function after 3 years with no maintenance?

- If the chance of them failing is 0.015, then the complement represents the chance of them being successful, so 1-0.015 = 0.985
- Since each light works independently, we may multiply the probabilities.
- We would take, formally:
- P(1st light works AND 2nd light works AND 3rd light works AND 4th light works…AND 25th light works)
- This is just 0.985 multiplied by itself 25 times or 0.985 to the 25th power or a 68.53% chance.

- Government data shows that 26% of the civilian labor force have at least 4 years of college and that 16% of the labor force work as laborers or operators of machines or vehicles. Can you conclude that because (0.26)(0.16)=0.0416, or about 4% of the labor force are college-educated laborers or operators? Explain.

- Absolutely not. In order to multiply probabilities, they must be independent. It is rather likely that whether or not you have a college education impacts whether or not you are a laborer/operator.

- This concludes this presentation.