CS107 hw2

Homework 2

page 44. 1) Counting from 0 to 999,999 yields 1,000,000 numbers.

Those which contain the digit 5 occur every 5th number: 0, 5, 10… Hence: $\frac{1,000,000}{5} = 200,000$ 6 digit numbers contain the digit 5.

3) Yes.

To decide this, we need to find the total number of possible initials. There are 26 letters in the English alphabet, hence, the number of possible initials ($n$) is: $n = 26^3 = 17,576$

Since there are only 17,576 possible sets of 3 initials, and 20,000 people live in the town, there must be at least two people with the same initials.

page 45.

18) The problem is asking for:

The count of different numbers of 4 digits, with no repeating digits. This is equal to:

$_{10} P_{4} = \frac{10!}{6!} = 5040$. Of those:

a) which do not start with a 0. This is equal to: $_{10} P_{4} - _{10} P_{3}= 5040 - 720 = 4320$

b) how many do not start with 01. This is equal to: $_{10} P_{4} - _{10} P_{2} = 5040 - 90 = 4950$

page 52.

16) The probability that a boy is sitting on a given chair (does not matter which one) is equal to:

$\frac{boys}{chairs} = \frac{12}{20} = \frac{3}{5}$

Where boys = number of boys, chairs = total number of chairs Note that this solution assumes the 20 individuals are all occupying chairs at the same time and thus the events of occupying a chair are mutually exclusive.

page 82.

2) We call event $E_1$ = soldier has antigen A, and event $E_2$ = soldier has blood type A. Then: $P(E_2 | E_1) =\frac{P(E_1 E_2)}{P(E_1)} = \frac{0.04}{0.04+0.41} = 0.0(8)$

page 87.

1) We call event $E_1$ = Robert commits perjury, and event $E_2$ = Susan has committed a crime. We want to calculate $P(E_1 | E_2)$: $P(E_1 | E_2) = \frac{P(E_1 \cdot E_2}{P(E_2)} = \frac{0.25 \cdot 0.65}{0.65} = 0.25$

4) 8 defective, 12 nondefective items are inspected without replacement. We call events $A_i$ = the i-th item inspected is defective, and events $B_i$ = the i-th item inspected is nondefective.

a) event $E_1$ = first four items inspected are defective. $P(E_1) = P(A_1 A_2 A_3 A_4) = P(A_1) \cdot P(A_2|A_1) \cdot P(A_3|A_1 A_2) \cdot P(A_4 | A_1 A_2 A_3)$

To calculate the above probability, it would help us to calculate $P(A_1 A_2), P(A_1 A_2 A_3)$. We already know $P(A_1)$, so we can use that to calculate the others. Hence: $P(A_1 A_2) = P(A_1) \cdot P(A_2|A_1) = \frac{8}{20} \cdot \frac{7}{19} \approx 0.147$ $P(A_1 A_2 A_3) = P(A_1) \cdot P(A_2 | A_1) \cdot P(A_3 | A_1 A_2) = \frac{8}{20} \cdot \frac{7}{19} \cdot \frac{6}{18} \approx 0.082$

The reasoning behind the result of $P(A_i | A_1 .. A_{i-1})$ is this: “What is the probability of selecting a(nother) defective item, given that I have already selected $i-1$ such items. Because of this dependency, we know the events that have to occur before this one, and what the numbers for defective and nondefective items are. Therefore: $P(A_1 A_2 A_3 A_4) = P(A_1) \cdot P(A_2 | A_1) \cdot P(A_3 | A_1 A_2) \cdot P(A_4 | A_1 A_2 A_3)= \frac{8}{20} \cdot \frac{7}{19} \cdot \frac{6}{18} \cdot \frac{5}{17} = \approx 0.0241$ Therefore, $P(E_1) \approx 0.0241$

b) Event $E_2$ = from the first three items, at least two are defective. The possible results can be split into:

  • $E_{21}$ = items 1 and 2 are defective $P(A_1 A_2) = P(A_1) \cdot P(A_2|A_1) = \frac{8}{20} \cdot \frac{7}{19} \approx 0.147$
  • $E_{22}$ = items 1 and 3 are defective, and item 2 is not (or we would have first case) $P(A_1 B_2 A_3) = P(A_1) \cdot P(B_2|A_1) \cdot P(A_3 | A_1 B_2) = \frac{8}{20} \cdot \frac{12}{19} \cdot \frac{7}{18} \approx 0.0982$
  • $E_{23}$ = items 2 and 3 are defective, and item 1 is not (or we would have first case) $P(B_1 A_2 A_3) = P(B_1) \cdot P(A_2|B_1) \cdot P(A_3|B_1 A_2) = \frac{12}{20} \cdot \frac{8}{19} \cdot \frac{7}{18} \approx 0.0982$

Therefore, $P(E_2) \approx 0.3435$

7) It does not matter. Some simple calculations for the case of 3 instead of 200 guests show that the probability of drawing the “you lose” paper is the same, no matter which order you choose.

For example, assume there are 3 guests, and 3 papers, one of which says “you lose.” The probability of drawing the “you lose” paper 1st, 2nd, or 3rd are:

$P(E_1) = \frac{1}{3}$ $P(E_2) = \frac{2}{3} \cdot \frac{1}{2}$, which is the probability of first guest drawing blank and me drawing “you lose.” $P(E_1) = \frac{2}{3} \cdot \frac{1}{2} \cdot 1$, which is the probability of first two guests drawing blank and me drawing “you lose.” (since that is all that is left, we have a certainty)

For the case of 200, clearly, drawing 1st yields a probability of losing of $\frac{1}{200}$. Drawing 2nd: $P(E_2) = \frac{199}{200} \cdot \frac{1}{199} = \frac{1}{200}$. By the following formula, we can show that the probability is the same for drawing “you lose” regardless of which order we choose:

$P(E_i) = (\prod_{j=1}^{i-1} \frac{200-j}{200-j+1}) \cdot \frac{1}{200-i+1}$. Expanding this product reduces the formula to: $P(E_i) = \frac{1}{200}$

page 97.

8) Let $G_i$ be the probability of hitting the target with gun i (i goes from 1 to 6). The probability of hitting a target by selecting a gun at random is: $P(E_1) = \sum_{i=1}^6 (\frac{1}{6} \cdot P(G_i)) = \frac{1}{6} \cdot (0.6 + 0.5 + 0.7 + 0.9 + 0.7 + 0.8) = 0.7$

11) Yes, the chances of getting a gold coin are equal for all participants. Intuitively, performing this experiment enough times yields equal probability for any person to receive a gold coin. More theoretically, we have shown in excercise 7 from page 87 (see above), that the order of drawing does not matter. This excercise is the same, except instead of drawing from 200 papers out of which 1 is distinct, we draw from 5 coins out of which 3 are distinct. Hence, the probability of each participant to get a gold coin is equal to 3/5.

page 106.

3) We will look at the probability that the suspect is guilty, given the new evidence that the criminal is (likely) left handed.

$P(G|L) = \frac{P(L|G) \cdot P(G)}{P(L|G) \cdot P(G) + P(L|I) \cdot P(I)}$, where L = criminal is left handed, G = suspect is guilty, I = suspect is innocent. Therefore:

$P(G|L) = \frac{0.23 \cdot 0.65}{0.23 \cdot 0.65 + 0.85 \cdot 0.35} = \frac{0.1495}{0.447} = 0.33$

As we suspect, since only 23% of people are left-handed and a witness claims criminal is likely (85%) left handed, our suspect is less likely to be guilty.

7) We look at the probability that a person has the cancer, given a positive test result.

$P(C|T) = \frac{P(T|C) \cdot P(C)}{P(T|C) \cdot P(C) + P(T|H) \cdot P(H)}$, where C = person has the cancer, T = test result is positive, H = the person is healthy (does not suffer from the cancer). Therefore:

$P(C|T) = \frac{0.92 \cdot 0.0002}{0.92 \cdot 0.0002 + 0.002 \cdot 0.9998} = \frac{0.000184}{0.0021836} = 0.0843$

9) We look at the probability that the collector buys a piece that she thinks is original, but she is mistaken; in other terms, the probability that the piece is not original, given that she thinks it is.

$P(F|T) = \frac{P(T|F) \cdot P(F)}{P(T|F) \cdot P(F) + P(T|O) \cdot P(O)}$, where O = the painting is an original, F = the painting is a fake, T = the collector thinks the painting is original

$P(F|T) = \frac{0.15 \cdot 0.25}{0.15 \cdot 0.25 + 0.85 \cdot 0.75} = \frac{0.0375}{0.675} = 0.0(5)$

page 119.

8) No, Cardano was not correct. Since the event A had probability $P(A) = \frac{3}{4}$, the occurence of that event in two consecutive independent experiments was $P(A A) = P(A) \cdot P(A) = \frac{3}{4} \cdot \frac{3}{4} = \frac{9}{16}$. So, the odds in favor were 9 to 7.

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