Chapter39 - 39.1. Model: The sum of the probabilities of...

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39.1. Model: The sum of the probabilities of all possible outcomes must equal 1 (100%). Solve: The sum of the probabilities is P A + P B + P C + P D = 1. Hence, 0.40 + 0.30 + P C + P D = 1 P C + P D = 0.30 Because P C = 2 P D , 2 P D + P D = 0.30. This means P D = 0.10 and P C = 0.20. Thus, the probabilities of outcomes C and D are 20% and 10%, respectively.
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39.2. Model: The probability that the outcome will be A or B is the sum of P A and P B . Solve: (a) Coin A Coin B Coin C HHH HHT HTH HTT THH THT TTH TTT (b) From the above table, we see that 2 heads and 1 tail occur three times (HHT, HTH, THH). Out of the possible eight outcomes, each outcome is equally probable and has a probability of occurrence of 1/8. So, the probability of getting 2 heads and 1 tail is 3/8 = 37.5%. (c) From the table, we see that at least two heads occur 4 times (HHH, HHT, HTH and THH). So, the probability of getting at least two heads is 4/8 = 50%.
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39.3. Model: The probability that the outcome will be A or B is the sum of P A and P B . Solve: (a) A regular deck of cards has 52 cards. Drawing a card from this deck has a probability of 1/52. Because there are 4 aces in the deck, the probability of drawing an ace is 4/52 = 0.077 = 7.7%. (b) Because there are 13 spades, the probability of drawing a spade is 13/52 = 0.25 = 25%.
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39.4. Model: The probability that the outcome will be A or B is the sum of P A and P B . The expected value is your best possible prediction of the outcome of an experiment. Solve: For each deck, there are 12 picture cards (4 Jacks, 4 Queens and 4 Kings). Because the probability of drawing one card out of 52 cards is 1/52, the probability of drawing a card that is a picture card is 12/52 = 23.1%. The number of picture cards that will be drawn is 0.231 × 1000 = 231.
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39.5. Model: The probability that the outcome will be A or B is the sum of P A and P B . Solve: (a) Each die has six faces and the faces have dots numbering from 1 to 6. We have two dice A and B. The various possible outcomes of rolling two dice are given in the following table. ABABAB 1 1 1 1 1 1 2 2 2 2 2 2 1 2 3 4 5 6 1 2 3 4 5 6 3 3 3 3 3 3 4 4 4 4 4 4 1 2 3 4 5 6 1 2 3 4 5 6 5 5 5 5 5 5 6 6 6 6 6 6 1 2 3 4 5 6 1 2 3 4 5 6 There are 36 possible outcomes. From the table, we find that there are six ways of rolling a 7 (1 and 6, 2 and 5, 3 and 4, 4 and 3, 5 and 2, 6 and 1). The probability is (1/36) × 6 = 1/6. (b) Likewise, the probability of rolling a double is 1/6. (c) There are 10 ways of rolling a 6 or an 8. The probability is (1/36) × 10 = 5/18.
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39.6. Model: The probability density of finding a photon is directly proportional to the square of the light- wave amplitude Ax () 2 . Solve: The probability of finding a photon within a narrow region of width δ x at position x is Prob(in x at x ) ∝ () x 2 Prob in at Prob in at xx x x 11 22 1 2 2 2 = Let N be the total number of photons and N 2 the number of photons detected at x 2 in a width x . The above equation simplifies to 2000 / N 10 V / m 0.10 mm 30 V / m 0.10 mm 30 V / m 10 V / m 18,000 NN N 2 2 2 2 2 2 2000 = ⇒= ( ) =
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39.7.
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Chapter39 - 39.1. Model: The sum of the probabilities of...

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