Intersections, Conditional Probability and Independence by Klara Grodzinsky

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1 Intersections, Conditional Probability and Independence by Klara Grodzinsky Previously, you have learned the additive rule for probabilities: P A B) =PA)+PB) PA B). If we solve for P A B), we have the expression P A B) =PA)+PB) PA B). When do we use this formula to find the probability of the intersection? Here s one example: Example: There is a 0% chance of rain today, a 50% chance of rain tomorrow, and a 5% chance of rain either today or tomorrow. What is the probability that it will rain today and tomorrow? Solution: LetAbe the event that it rains today and B the event that it rains tomorrow. We wish to find the probability of A AND B, orpa B). We are given P A) =0., P B) =0.5, and P A B) =0.5. Now just plug the numbers into the formula: P A B) =PA)+PB) PA B)= = 0.45, so there is a 45% chance of rain on both days. What happens to our above formula if A and B are mutually exclusive? Recall that mutually exclusive means P A B) = 0; i.e., the events A and B have an empty intersection. Be VERY careful not to confuse mutually exclusive events with independent events which we will discuss below). CONDITIONAL PROBABILITY Let s start with an example: suppose a family has two children, where the sample space looks at each child as girl G) or boy B). The sample space for this problem is S = {GG, GB, BG, BB}. We will also assume that the probability of having a baby boy is 0.5. Now suppose we wish to find the probability that the family has one boy and one girl, but we also have the information that at least one of the children is a boy. What is this probability? 1

2 Solution: One way to think about the problem is the following: since we know at least one child is a boy, we can rewrite our sample space as {GB, BG, BB}. Now since each outcome in our sample space is equally likely, we can compute that the probability of one boyandonegirlisp= 2. Another way to obtain the same answer is to notice that 2 = 2/4 /4 = P GB)+PBG) PB) Pgirl & boy) = P at least 1 boy) P [girl & boy] [at least 1 boy]) =. P at least 1 boy) Now let s introduce some notation. Let E and F be two events. If we wish to find the probability that event E occurs given that event F occurs in other words, the probability that E happens if we already know F happened then we denote this probability by P E F ). We read this as the probability of E given F, or the conditional probability. We can generalize the above example to the following formula for conditional probability: P E F )= PE F) PF), PF E)= PE F). PE) Written slightly differently, we have the product rule for probabilities: P E F )=PE F) PF)=PF E) PE). Note that we now have yet another way of computing the probability of the intersection of two events. However, we will ONLY use this formula when either P E F )orpf E) is easy to compute. Example: An experiment consists of drawing two cards, without replacement, from a deck of 52. Find the probability that the first card is red and the second card is a spade. Solution: Let E be the event that the first card is red, and F the event that the second card is a spade. Then we are looking for P E F ) watch the wording: the problem asks for the first card to be red AND the second card to be a spade). If we were to attempt this problem using the additive rule, we would have to find P E F ). But finding 2

3 the probability that the first card is red OR the second card is a spade is actually quite difficult. Let s try using conditional probability instead. In this case, we can easily compute P E) andpf E). First, P E) = = 1 2 since half of the deck is red. Once we have drawn out a red card, we have 51 cards remaining, 1 of which are spades. Thus, the probability that we draw a spade second is P F E) = 1 51.WethencalculatePE F)by using the formula P E F )=PF E) PE)= = 1 2. INDEPENDENCE Two events E and F are independent if P E F )=PE)andPF E)=PF). BE CAREFUL not to confuse independent events with mutually exclusive events! If the events E and F have nothing in common, then they are mutually exclusive. But if the results of one event do not affect the results of the other event, they are independent. For example, suppose we toss a fair coin twice. Then the events E = {thefirsttossis ahead},f ={thefirsttossisatail}are mutually exclusive, sincebotheand F cannot occur at the same time. However, if G = {the second toss is a head}, then the events E and G are independent, since a head on the first toss does not change the probability of a head on the second toss. Likewise, the events F and G are independent. For independent events, the product rule simply becomes P E F )=PE) PF). The above formula can also be used as a test for independence. BE CAREFUL not to use this formula unless you already KNOW that the events are independent! Since, in the above example, E and G are independent, we can compute P HH)=PE G)=PE) PG)= =1 4. Example: Draw out two cards from a deck of 52. What is the probability that the second card is a 5 if the cards are drawn a) without replacement or b) with replacement? Solution: a) When we draw the cards without replacement, we use the conditional probability. WHY? Because the probability of drawing out a 5 CHANGES after we remove a

4 card from the deck, so the first card and second card are NOT independent of one another. Our probabilities also change depending on whether or not the first card was a 5. Keep in mind that once we draw out a card, there are only 51 remaining cards. If the first card is a 5, then we have only three 5 s left in the deck; if the first card is not a 5, we have all four 5 s left in the deck. So we have: P 2 nd = 5 )=P1 st = 5 2 nd = 5 )+P1 st 5 2 nd = 5 ) =P1 st = 5 ) P2 nd = 5 1 st = 5 )+P1 st 5 ) P2 nd = 5 1 st 5 ) = = 1 1. b) When we draw the cards with replacement, the first and second cards are now independent. WHY? After we draw the first card, we place it back into the deck, so the probability of drawing any given card on the second draw is the same as it was on the first draw. Now it does not matter which card we drew out first, since all four 5 s are still in the deck, and all 52 cards are also in the deck. So P 2 nd = 5 )= 4 52 = 1 1. Example: An experiment consists of tossing a coin times. The coin is biased so that heads is twice as likely to occur as tails. Find the probability that the coin lands on heads exactly times if it is known that at least two heads appear. Solution: Watch for wording: if it is known is the key phrase that tells us to use conditional probability. Let s set up our events: let E be the event that the coin lands on heads exactly times, and let F be the event that at least two heads appear. We are asked to find the probability P E F ). Since heads is twice as likely as tails, P H) = 2 and P T )= 1. Now, we must find P E F )andpf). The easiest way to find P E F ) in this case is to just think about what the intersection means. If we have exactly heads AND at least two heads, then we must have exactly heads or we would not satisfy both conditions). In other words, P E F )=PE) for this particular problem. To calculate P E), we use the fact that each toss of the coin is independent of the previous toss. So, if) we wish to have exactly heads on tosses, we first pick the slots for heads in ways. The probability of having a head on those tosses is 2,and 4

5 the probability of a tail on the remaining tosses is 1. As each toss is independent, we can find the probability of heads AND tails by simply multiplying all the probabilities: P E) = ) ) ) 2 1. To find P F ), we use the fact that at least two heads is the complement of zero heads or one head. If we have zero heads, then all tosses are tails; if we have one head, we select a slot for that head, and the remaining 9 tosses are tails. Thus, the probability is: [ 1 ) P F )=1 + 1 Then we can find the conditional probability as follows: P E F )= PE F) PF) = PE) PF) = 1 1 ) ) 1 ) ] ) 1 ) 2 1 ) ) ) ) 9 ) EXERCISES 1. An urn contains red balls, white balls, and 8 yellow balls. Three balls are drawn out, without replacement, and placed in positions 1, 2, and. Find the probability that: a) the first two balls are yellow and the third is red. b) two balls are yellow and one is red, in any order. c) the third ball is yellow if it is known that the first ball was yellow and the second was white. 2. Repeat problem #1 if the balls are drawn with replacement. 5

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