40 1 Model The sum of the probabilities of all possible outcomes must equal 1 100 Solve The sum of the probabilities is PA PB PC PD 1 Hence 0 40 0 30 PC PD 1 PC PD 0 30 Because PC 2PD 2PD PD 0 30 This means PD 0 10 and PC 0 20 Thus the probabilities of outcomes C and D are 20 and 10 respectively 40 2 Model The probability that the outcome will be A or B is the sum of PA and PB Solve a Coin A Coin B Coin C H H H H T T T T H H T T H H T T H T H T H T H T 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 a A regular deck of cards has 52 cards Drawing a card from this deck has a probability of 1 52 40 3 Model The probability that the outcome will be A or B is the sum of PA and PB Solve 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 40 4 Model The probability that the outcome will be A or B is the sum of PA and PB 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 40 5 Model The probability that the outcome will be A or B is the sum of PA and PB Solve The various possible outcomes of rolling two dice are given in the following table a Each die has six faces and the faces have dots numbering from 1 to 6 We have two dice A and B A 1 1 1 1 1 1 2 2 2 2 2 2 B 1 2 3 4 5 6 1 2 3 4 5 6 A 3 3 3 3 3 3 4 4 4 4 4 4 B 1 2 3 4 5 6 1 2 3 4 5 6 A 5 5 5 5 5 5 6 6 6 6 6 6 B 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 40 6 Model The probability density of finding a photon is directly proportional to the square of the light wave amplitude 2 A x Solve The probability of finding a photon within a narrow region of width x at position x is Prob in x at x 2 A x x Prob in Prob in x 1 x 2 at at x 1 x 2 2 A x 1 x 2 A x 2 x Let N be the total number of photons and N2 the number of photons detected at x2 in a width x The above equation simplifies to 2000 N N N 2 2 10 V m 0 10 mm 2 30 V m 0 10 mm N 2 2000 30 V m 10 V m 2 2 18 000 40 7 Visualize Combine Equations 40 10 and 40 11 to show that N is proportional to A x 2 x A x 2 A x 1 2 2 x 2 x 1 N in x at 2 N in x at 1 x 2 N tot x 1 N tot We are given N 1 6000 x 1 0 10 mm A x 1 200 V m N 2 3000 and x 2 0 20 mm We are not given Ntot but it cancels anyway Solve Solve the above equation for A x 2 A x 2 A x 1 x N 1 x N 2 in in x 2 x 1 at x 2 at x 1 200 V m 0 10 mm 3000 0 20 mm 6000 100 V m Assess The answer is half of the wave amplitude at the other strip which seems reasonable 40 8 Solve The probability that a photon arrives at this 0 10 mm wide strip is Prob in 0 10 mm at x N 1 0 10 10 P x x where N is the number of photons detected in the strip and the total number of photons is 1 0 1010 We have 20 m 0 10 10 m N 2 0 107 1 0 10 N 3 1 10 40 9 Model See Example 40 1 Visualize We are given tot 9 2 0 10 5 0 10 photon goes through the slit is Solve Solve for the probability density P x in Equation 40 12 9 at 2 0 10 12 N 12 5 0 10 Prob in x x at x x in N 0 00040 and x 0 10 mm The probability that a P x Prob in x x at x 0 00040 4 1 0 10 m 4 0 m 1 Assess This result is similar to that in Example 40 1 is a probability which is dimensionless The units of x are m so the units of 40 10 Solve 2 x x are m 1 and thus the units of are m 1 2 2 x 40 11 Model The probability of finding a particle at position x is determined by x 2 Solve a The probability of detecting an electron is Prob in x at x At x 0 mm the number of 2 x x electrons landing is calculated as follows N 1 0 10 6 0 mm 2 x N mm 0 010 mm 1 0 10 3333 1 3 1 b Likewise the number of electrons landing at x 2 0 mm is 2 0 mm xN N total 1 2 0 111 mm 0 010 mm 1 0 10 1111 6 6 40 12 Visualize Solve a We assume that at some point in between each of the peaks …
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