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Ch. 19- Probability Day 1

Ch. 19- Probability Day 1. Ex. 1 Pat keeps records on how often she fills her car with petrol. The table shows the frequencies of the number of days between refills . Estimate the likelihood that: a) there is a four day gap between refills

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Ch. 19- Probability Day 1

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  1. Ch. 19- Probability Day 1

  2. Ex. 1 Pat keeps records on how often she fills her car with petrol. The table shows the frequencies of the number of days between refills. Estimate the likelihood that: a) there is a four day gap between refills b) there is at least a four day gap between refills. She refilled the car a total of 190 times. There was a four day gap between refills a total of 17 refills. a) Probability = 17/190 At least a four-day gap includes, 4,5, or 6 day gaps. So: b) Probability = (17+6+1)/190 = 24/190

  3. A sample space is the set of all possible outcomes of an experiment. Ex. 2 List the sample space of possible outcomes for: a) tossing a coin b) rolling a die a) Two possible outcomes: b) 6 possible outcomes: Ex. 3 Illustrate the possible outcomes when 2 coins are tossed by using a 2-dimensional grid: coin 1 H T H T coin 2 Each of these points represents one of the possible outcomes in the sample space of tossing two coins:

  4. Ex. 4 Illustrate, using a tree diagram, the possible outcomes when: • tossing two coins • drawing two marbles from a bag, containing several red, green, and • yellow marbles coin 1 coin 2 Each branch represents a possible outcome: H T H T a) H T marble 1 marble 2 R G Y R G Y We could repeat these branches as many times as needed for marbles 3, 4, etc… b) R G Y Tree diagrams are very important and will be used often in solving probability problems! R G Y

  5. Ex. 5 A ticket is randomlyselected from a basket containing 3 green, • 4 yellow, and 5 blue tickets. Determine the probability of getting: • a green ticket b) a green or yellow ticket • c) an orange ticket d) a green, yellow, or blue ticket The sample space is: = 12 outcomes a) P(G) = 3/12 = 1/4 b) P(G or Y) = 7/12 c) P(O) = 0 d) P(G,Y, or B) = 1

  6. In the last problem, an orange ticket had a probability of 0. The probability of getting a green, yellow, or blue ticket was 1. For any event E, Ex. 6 An ordinary 6-sided die is rolled once. Determine the chance of: • getting a 6 b) not getting a 6 • c) getting a 1 or 2 d) not getting a 1 or 2 a) P(6) = 1/6 b) P(not 6) = 5/6 These are complimentary events. Notice that: P(6) + P(not 6) = 1 c) P(1 or 2) = 2/6 = 1/3 d) P(not 1 or 2) = 4/6 = 2/3

  7. If E is an event, then is the complimentary event of E. P(E not occurring) = 1- P(E occurring)

  8. Ex. 7 Use a two-dimensional grid to illustrate the sample space for tossing a coin and rolling a die simultaneously. From this grid determine the probability of: a) tossing a head b) getting a tail and a 5 c) getting a tail or a 5 coin H T From the grid, you can see there are 12 possible outcomes. 1 2 3 4 5 6 die a) P(H) = 6/12 = 1/2 b) P(T and 5) = 1/12 6 chances of getting a tail (including 5) + 1 chance of getting a 5, that’s not a tail c) P(T or 5) = = 7/12

  9. Independent events are events where the occurrence of one event does not affect the occurrence of the other event. If A and B are independent events, then This can be extended as needed: Dependent events are events where the occurrence of one of the events does affect the occurrence of the other event. If A and B are dependent events, then (More to come on dependent events later…)

  10. Ex. 8 A coin and a die are tossed simultaneously. Determine the probability of getting a head and a 3 without using a grid. First note that these are independent events. P(head and 3) = P(H) x P(3)

  11. Ex. 9 A box contains 4 red and 2 yellow tickets. Two tickets are randomly selected, one by one from the box, without replacement. Find the probability that: a) both are red b) the first is red and the second is yellow a) P(both red) = P(first red and second red) =P(first red) x P(second red) b) P(first red and second yellow) = P(first red) x P(second yellow)

  12. Ex. 10 A hat contains tickets with numbers 1,2,3,…,19,20 printed on them. If 3 tickets are drawn from the hat, without replacement, determine the probability that all are prime numbers. Of the 20 numbers, 8 are prime. We want P(3 primes). =P(prime on 1st draw and prime on 2nd draw and prime on 3rd draw) =P(prime on 1st draw) x P(prime on 2nd draw) x P(prime on 3rd draw)

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