12.6. Or and And Problems


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1 12.6 Or and And Problems
2 Or Problems P(A or B) = P(A) + P(B) P(A and B) Example: Each of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 is written on a separate piece of paper. The 10 pieces of paper are then placed in a bowl and one is randomly selected. Find the probability that the piece of paper selected contains an even number or a number greater than 5.
3 Solution P(A or B) = P(A) + P(B) P(A and B) even or even and P = P(even) + P(greater 5) greater than 5 greater than = = 10 Thus, the probability of selecting an even number or a number greater than 5 is 7/10.
4 Example Each of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 is written on a separate piece of paper. The 10 pieces of paper are then placed in a bowl and one is randomly selected. Find the probability that the piece of paper selected contains a number less than 3 or a number greater than 7.
5 Solution 2 P (less than 3) = 10 P (greater than 7) = 3 10 There are no numbers that are both less than 3 and greater than 7. Therefore, = =
6 Mutually Exclusive Two events A and B are mutually exclusive if it is impossible for both events to occur simultaneously.
7 Example One card is selected from a standard deck of playing cards. Determine the probability of the following events. a) selecting a 3 or a jack b) selecting a jack or a heart c) selecting a picture card or a red card d) selecting a red card or a black card
8 Solutions a) 3 or a jack 4 4 P(3) + P(jack) = = = b) jack or a heart jack and P(jack) + P(heart) P = + heart = = 52 13
9 Solutions continued c) picture card or red card picture & P(picture) + P(red) P = + red card = = d) red card or black card P(red) + P(black) = = = 1 52
10 Independent Events Event A and Event B are independent events if the occurrence of either event in no way affects the probability of the occurrence of the other event. Experiments done with replacement will result in independent events, and those done without replacement will result in dependent events.
11 And Problems If A and B are independent, P(A and B) = P(A) P(B) Example: Two cards are to be selected with replacement from a deck of cards. Find the probability that two red cards will P( A) P( B) = P( red) P( red) be selected = = = 2 2 4
12 Example Two cards are to be selected without replacement from a deck of cards. Find the probability that two red cards will be selected. PA ( ) PB ( ) = Pred ( ) Pred ( ) = = =
13 Example A package of 30 tulip bulbs contains 14 bulbs for red flowers, 10 for yellow flowers, and 6 for pink flowers. Three bulbs are randomly selected and planted. Find the probability of each of the following. All three bulbs will produce pink flowers. The first bulb selected will produce a red flower, the second will produce a yellow flower and the third will produce a red flower. None of the bulbs will produce a yellow flower. At least one will produce yellow flowers.
14 Solution 30 tulip bulbs, 14 bulbs for red flowers, 10 for yellow flowers, and 6 for pink flowers. All three bulbs will produce pink flowers. P( 3 pink ) = P(pink 1) P(pink 2) P(pink 3) = = 1 203
15 Solution 30 tulip bulbs, 14 bulbs for red flowers, 0010 for yellow flowers, and 6 for pink flowers. The first bulb selected will produce a red flower, the second will produce a yellow flower and the third will produce a red flower. ( ) P red,yellow,red = P(red) P(yellow) P(red) = =
16 Solution 30 tulip bulbs, 14 bulbs for red flowers, 0010 for yellow flowers, and 6 for pink flowers. None of the bulbs will produce a yellow flower. first not second not third not P( none yellow ) = P P P yellow yellow yellow = =
17 Solution 30 tulip bulbs, 14 bulbs for red flowers, 0010 for yellow flowers, and 6 for pink flowers. At least one will produce yellow flowers. P(at least one yellow) = 1 P(no yellow) 57 = =
18 12.7 Conditional Probability
19 Conditional Probability In general, the probability of event E 2 occurring, given that an event E 1 has happened (or will happen; the relationship does not matter) is called conditional probability and is written P(E 2 E 1 ).
20 Example Given a family of two children, and assuming that boys and girls are equally likely, find the probability that the family has a) two girls. b) two girls if you know that at least one of the children is a girl. c) two girls given that the older child is a girl.
21 Solutions a) two girls There are four possible outcomes BB, BG, GB, and GG. 1 P(2 girls) = 4 b) two girls if you know that at least one of the children is a girl 1 P(both girls at least one is a girl) = 3
22 Solutions continued two girls given that the older child is a girl 1 P(both girls older child is a girl) = 2
23 Conditional Probability For any two events, E 1, and E 2, PE ( E) 2 1 = ne ( and E ) 1 2 ne ( ) 1
24 Example Use the results of the taste test given at a local mall. If one person from the sample is selected at random, find the probability the person selected Prefers Peppermint Prefers Wintergreen Total Men Women Total
25 Example continued a) prefers peppermint P (peppermint) = b) is a woman P (woman) =
26 Example continued c) prefers peppermint, given that a woman is selected P (peppermint woman) = = d) is a man, given that the person prefers wintergreen 55 P (man wintergreen) = 127
27 Next Steps Read all examples in the text for the corresponding sections. Homework problems from text: 12.6,p.705: 2743,odds; 6366, all 12.7, p. 710: 523, odds;4752, all; 6570, all Do Online Homework Do online quiz after finishing homework for both sections
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