Ch. 16 Probability

Question 1

Probability of event X =

Answer 1

Number of outcomes in which x occurs, divided by
Total number of outcomes in the experiment

Question 2

What is a sample space? What is the probability of a sample space?

Answer 2

Set of all possible outcome of the experiment
P(Sample Space) = 1

Question 3

What are complementary events?

Answer 3

Two events that share no common outcomes, but together cover every possible outcome.
If one occurs, the other doesn’t.
P(A) + P(A’) = 1

Question 4

How can you use complementary events to make calculations easier?

Answer 4

P(A) + P(A’) = 1
P(A) = 1 - P(A’) or P(A’) = 1 - P(A)

Figure out if it’s easier to calculate P(A) or P(A’), then just do 1 - P(..) to get the complementary event probability.
*MAKE SURE they’re truly complementary!! They truly cover ALL outcomes & share no common outcomes.

Question 5

Independent events in probability meaning

Answer 5

The fact that Event A occurs does not change the probability that Event B occurs

Question 6

What is the probability of event A, event B and event C all occurring if they are all independent of each other?

Answer 6

P(A and B and C) = P(A) × P(B) × P(C)

Question 7

If A & B are independent, what else is independent?

Answer 7

Their complements, A’ & B’ are also independent

Question 8

If A & B’ are independent, what else is independent?

Answer 8

Their complements, A’ & B are also independent

Question 9

Dependent events in probability meaning

Answer 9

If the probability of Event B changes because of the occurrence of Event A, they’re dependent.

Question 10

What is the probability of event A & event B occurring if they are dependent?

Answer 10

P(A and B) = P(A) × P(B|A)

Question 11

What does it mean for two events to be mutually exclusive?

Answer 11

They can’t occur together at the same time.

For example, getting heads and getting tails in one coin flip is impossible, so the two events are mutually exclusive.

Question 12

When flipping a fair coin 3 times, what is the probability of getting heads first, then heads, then tails?

Answer 12

These are independent events, which means:
P(A and B and C) = P(A) × P(B) × P(C)
P(HHT) = 1/2 × 1/2 × 1/2 = 1/8

Question 13

If two events are mutually exclusive, what is the formula for the probability that event A happens OR event B happens?

Answer 13

P(A or B) = P(A) + P(B)

Can be extended: P(A or B or C or D…) = P(A) + P(B) + P(C) + P(D) +…

Question 14

If two events are NOT mutually exclusive, what is the formula for the probability that event A happens OR event B happens?

Answer 14

P(A or B) = P(A) + P(B) - P(A and B)
Need to remove overlap/ co-occurrences, otherwise you’re overstating the probability.

Question 15

What is the traditional way to solve for multiple outcomes?

Answer 15

To solve:
Step 1: Convert events to letters, use permutations to determine the number of outcomes producing the event
Step 2: Determine the probability for one event
Step 3: P = the number of outcomes producing the event x probability of one outcome

Question 16

What does it mean when a problem has multiple outcomes? How can you tell? (e.g., a coin is flipped 4 times, what is the probability that the coin gets heads exactly 2 times?)

Answer 16

This is a multiple outcomes question because there are MANY outcomes that will satisfy 2 heads, 2 tails: HHTT, HTHT, TTHH, etc.

Question 17

How do you solve “at least” questions? e.g., a coin is tossed 3 times, what is the probability it lands on heads at least 2 times?

Answer 17

At least 2 times means it lands on heads 2 or 3 times.
You have to break it out into all possible scenarios, then find probability for each outcome, then add the probabilities.
Scenario 1: It lands on heads 2 times
Scenario 2: It lands on heads 3 times

If it lands on heads 2 times, this is multiple outcomes: # outcomes x probability of one outcome
If it lands on heads 3 times, P(HHH) = 1/2 x 1/2 x 1/2

Add the two probabilities together because the two scenarios are mutually exclusive. Either scenario 1 OR scenario 2 is happening, can’t be both.

Question 18

Is there a shortcut for solving “at least 1” problems?

Answer 18

Yes, use complementary events - make sure: 1. they don’t share ANY common outcomes and 2. together they cover every possible outcome.
P(A) + P(A’) = 1
P(at least 1) + P(none) = 1
P(at least 1) = 1 - P(none)

It’s usually easier to find P(none)

Question 19

What is the combinatorics way to solve for multiple outcomes?

Answer 19

The formula is # of outcomes in which the items can be selected / # of total outcomes or groups that can be created

Use combinations to find both numerator and denominator.

Question 20

How would you set this up to solve the combinatorics way? There are 4 even and 7 odd numbered cards. If Giselle were to randomly select 4 cards, what is the probability that she would select 2 even numbered cards and 2 odd numbered cards?

Answer 20

Probability = # favorable / # total

For # favorable, we need 2 out of 4 even and 2 out of 7 odd. We need to find the number of groups of 2 even cards & 2 odd cards:
4C2 x 7C2

For # total, we need find the total number of ways to select 4 cards out of 11 cards: 11C4

Answer: [ 4C2 x 7C2 ] / 11C4 = (6 x 21 ) / 330 = 21/55

Question 21

What is the strategy when asked the probability of some event occuring where certain items MUST get selected?

Answer 21

Similar to combinatorics, pretend the two items have already been selected - for the numerator, find the total number of groups with the two items already selected; for the denominator, find the total number of groups:

The formula is # of ways that some number of items must be selected / # of ways that ALL items must be selected