13) when a pair of dice are rolled there are 36 different possible outcomes: 1–1, 1–2, ... 6–6. if a pair of dice are rolled 3 times, what is the probability of getting a sum of 7 every time? round to eight decimal places.
a) 0.0476 b) 0.00462963 c) 0.00735168 d) 0.00291545

14) in a homicide case 4 different witnesses picked the same man from a line up. the line up contained 5 men. if the identifications were made by random guesses, find the probability that all 4 witnesses would pick the same person.
a) 0.0016 b) 0.8 c) 0.0009766 d) 0.008

15) you are dealt two cards successively (without replacement) from a shuffled deck of 52 playing cards. find the probability that the first card is a king and the second card is a queen. express your answer as a simplified fraction.
a) $\frac{2}{13}$ b) $\frac{1}{663}$ c) $\frac{4}{663}$ d) $\frac{13}{102}$

16) a irs auditor randomly selects 3 tax returns from 49 returns of which 7 contain errors. what is the probability that she selects none of those containing errors? round to four decimal places.
a) 0.0019 b) 0.6231 c) 0.6297 d) 0.0029

17) among the contestants in a competition are 46 women and 23 men. if 5 winners are randomly selected, what is the probability that they are all men? round to five decimal places.
a) 0.13169 b) 0.03125 c) 0.02455 d) 0.00299

18) a sample of 4 different calculators is randomly selected from a group containing 47 that are defective and 29 that have no defects. what is the probability that all four of the calculators selected are defective? round to four decimal places.
a) 0.1449 b) 7.5098 c) 0.1463 d) 0.1390

19) the table below describes the smoking habits of a group of asthma sufferers.
\begin{tabular}{l|cccc}
& nonsmoker & light smoker & heavy smoker & total \\ \hline
men & 425 & 38 & 35 & 498 \\
women & 381 & 32 & 43 & 456 \\
total & 806 & 70 & 78 & 954 \\
\end{tabular}
if two different people are randomly selected from the 954 subjects, find the probability that they are both women. round to four decimal places.
a) 0.2282 b) 0.2285 c) 0.1595 d) 0.000004809

Question

13) when a pair of dice are rolled there are 36 different possible outcomes: 1–1, 1–2, ... 6–6. if a pair of dice are rolled 3 times, what is the probability of getting a sum of 7 every time? round to eight decimal places.
a) 0.0476  b) 0.00462963  c) 0.00735168  d) 0.00291545

14) in a homicide case 4 different witnesses picked the same man from a line up. the line up contained 5 men. if the identifications were made by random guesses, find the probability that all 4 witnesses would pick the same person.
a) 0.0016  b) 0.8  c) 0.0009766  d) 0.008

15) you are dealt two cards successively (without replacement) from a shuffled deck of 52 playing cards. find the probability that the first card is a king and the second card is a queen. express your answer as a simplified fraction.
a) $\frac{2}{13}$  b) $\frac{1}{663}$  c) $\frac{4}{663}$  d) $\frac{13}{102}$

16) a irs auditor randomly selects 3 tax returns from 49 returns of which 7 contain errors. what is the probability that she selects none of those containing errors? round to four decimal places.
a) 0.0019  b) 0.6231  c) 0.6297  d) 0.0029

17) among the contestants in a competition are 46 women and 23 men. if 5 winners are randomly selected, what is the probability that they are all men? round to five decimal places.
a) 0.13169  b) 0.03125  c) 0.02455  d) 0.00299

18) a sample of 4 different calculators is randomly selected from a group containing 47 that are defective and 29 that have no defects. what is the probability that all four of the calculators selected are defective? round to four decimal places.
a) 0.1449  b) 7.5098  c) 0.1463  d) 0.1390

19) the table below describes the smoking habits of a group of asthma sufferers.
\begin{tabular}{l|cccc}
& nonsmoker & light smoker & heavy smoker & total \\ \hline
men & 425 & 38 & 35 & 498 \\
women & 381 & 32 & 43 & 456 \\
total & 806 & 70 & 78 & 954 \\
\end{tabular}
if two different people are randomly selected from the 954 subjects, find the probability that they are both women. round to four decimal places.
a) 0.2282  b) 0.2285  c) 0.1595  d) 0.000004809

Accepted Answer

### Question 13
# Explanation:
## Step1: Find probability of sum 7 in one roll
The pairs that sum to 7 when rolling two dice are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). So there are 6 favorable outcomes. Total outcomes when rolling two dice: 36. Probability of sum 7 in one roll: $P(7) = \frac{6}{36} = \frac{1}{6}$.
## Step2: Probability of sum 7 three times (independent events)
Since the rolls are independent, the probability of getting sum 7 three times is $(\frac{1}{6})^3$. Calculate: $(\frac{1}{6})^3 = \frac{1}{216} \approx 0.00462963$.
# Answer: B) 0.00462963

### Question 14
# Explanation:
## Step1: Probability one witness picks a man
There are 5 men in the lineup. Probability one witness picks a specific man (the same one) by random guess: $\frac{1}{5}$.
## Step2: Probability all 4 witnesses pick the same man
Since the witnesses' guesses are independent, the probability all 4 pick the same man is $(\frac{1}{5})^3$? Wait, no. Wait, first witness picks any man (prob 1), then the next three must pick the same man as the first. So first witness: 1 (any man), second: $\frac{1}{5}$, third: $\frac{1}{5}$, fourth: $\frac{1}{5}$. So probability: $1 	imes \frac{1}{5} 	imes \frac{1}{5} 	imes \frac{1}{5} = \frac{1}{125} = 0.008$? Wait, no, wait. Wait, the problem is "the same man" (any specific man? Or any man, as long as all pick the same). Wait, if we consider that the first witness picks a man (say man A), then the next three must pick man A. The probability the first witness picks man A: $\frac{1}{5}$, then the second: $\frac{1}{5}$, third: $\frac{1}{5}$, fourth: $\frac{1}{5}$. But actually, the first witness can pick any of the 5 men, and then the others must pick that same man. So the number of favorable outcomes: 5 (one for each man: all pick man 1, all pick man 2, ..., all pick man 5). Total possible outcomes for 4 witnesses: $5^4$. So probability: $\frac{5}{5^4} = \frac{1}{5^3} = \frac{1}{125} = 0.008$. Wait, but option D is 0.008. Wait, but let's check again. Wait, the line up has 5 men. Each witness picks a man at random. We want the probability that all 4 pick the same man (could be any of the 5). So for a specific man (e.g., man 1), the probability all 4 pick man 1 is $(\frac{1}{5})^4$. There are 5 such men, so total probability is $5 	imes (\frac{1}{5})^4 = \frac{5}{625} = \frac{1}{125} = 0.008$. So answer is D) 0.008. Wait, but let's check the options. Option D is 0.008. So that's correct.
# Answer: D) 0.008

### Question 15
# Explanation:
## Step1: Probability first card is King
There are 4 Kings in a deck of 52 cards. So probability first card is King: $P(K_1) = \frac{4}{52} = \frac{1}{13}$.
## Step2: Probability second card is Queen (without replacement)
After drawing a King, there are 51 cards left, and 4 Queens. So probability second card is Queen: $P(Q_2|K_1) = \frac{4}{51}$.
## Step3: Multiply the probabilities (independent? No, dependent, so use multiplication rule for dependent events: $P(K_1 \cap Q_2) = P(K_1) 	imes P(Q_2|K_1)$
Calculate: $\frac{4}{52} 	imes \frac{4}{51} = \frac{16}{2652} = \frac{4}{663}$.
# Answer: C) $\frac{4}{663}$

### Question 16
# Explanation:
## Step1: Total tax returns and error-free returns
Total tax returns: 49. Error-free returns: $49 - 7 = 42$.
## Step2: Probability of selecting 3 error-free returns (without replacement)
We use combinations. The number of ways to choose 3 from 42 error-free: $C(42, 3)$. The number of ways to choose 3 from 49: $C(49, 3)$. Probability: $\frac{C(42, 3)}{C(49, 3)}$.
Calculate $C(n, k) = \frac{n!}{k!(n - k)!}$. So $C(42, 3) = \frac{42!}{3!39!} = \frac{42 	imes 41 	imes 40}{3 	imes 2 	imes 1} = 11480$. $C(49, 3) = \frac{49!}{3!46!} = \frac{49 	imes 48 	imes 47}{3 	imes 2 	imes 1} = 18424$. Then probability: $\frac{11480}{18424} \approx 0.6231$? Wait, no: 11480 ÷ 18424 ≈ 0.6231? Wait, 42×41×40=68880; 68880/(6)=11480. 49×48×47=108624; 108624/6=18104? Wait, wait, 49×48=2352; 2352×47=109544? Wait, no, 49×48=2352, 2352×47: 2352×40=94080, 2352×7=16464, total 94080+16464=110544. Then divide by 6: 110544/6=18424. Then 42×41×40=68880; 68880/6=11480. Then 11480/18424 ≈ 0.6231? Wait, 11480 ÷ 18424: let's divide numerator and denominator by 8: 1435/2303 ≈ 0.6231. So yes, approximately 0.6231. So answer is B) 0.6231.
# Answer: B) 0.6231

### Question 17
# Explanation:
## Step1: Total contestants and men
Total contestants: $46 + 23 = 69$. Number of men: 23.
## Step2: Probability of selecting 5 men (without replacement)
Use combinations. Number of ways to choose 5 men: $C(23, 5)$. Number of ways to choose 5 contestants: $C(69, 5)$. Probability: $\frac{C(23, 5)}{C(69, 5)}$.
Calculate $C(23, 5) = \frac{23!}{5!18!} = \frac{23 	imes 22 	imes 21 	imes 20 	imes 19}{5 	imes 4 	imes 3 	imes 2 	imes 1} = 33649$. $C(69, 5) = \frac{69!}{5!64!} = \frac{69 	imes 68 	imes 67 	imes 66 	imes 65}{5 	imes 4 	imes 3 	imes 2 	imes 1} = 11238513$. Then probability: $\frac{33649}{11238513} \approx 0.00299$. Wait, 33649 ÷ 11238513 ≈ 0.00299. So answer is D) 0.00299.
# Answer: D) 0.00299

### Question 18
# Explanation:
## Step1: Total calculators and defective calculators
Total calculators: $47 + 29 = 76$. Number of defective calculators: 47.
## Step2: Probability of selecting 4 defective calculators (without replacement)
Use combinations. Number of ways to choose 4 defective: $C(47, 4)$. Number of ways to choose 4 calculators: $C(76, 4)$. Probability: $\frac{C(47, 4)}{C(76, 4)}$.
Calculate $C(47, 4) = \frac{47!}{4!43!} = \frac{47 	imes 46 	imes 45 	imes 44}{4 	imes 3 	imes 2 	imes 1} = 178365$. $C(76, 4) = \frac{76!}{4!72!} = \frac{76 	imes 75 	imes 74 	imes 73}{4 	imes 3 	imes 2 	imes 1} = 1262340$. Then probability: $\frac{178365}{1262340} \approx 0.1413$? Wait, no, wait: 47×46=2162, 2162×45=97290, 97290×44=4280760; divide by 24: 4280760/24=178365. 76×75=5700, 5700×74=421800, 421800×73=30791400; divide by 24: 30791400/24=1282975? Wait, no, 76×75×74×73: 76×75=5700, 5700×74=421800, 421800×73: 421800×70=29526000, 421800×3=1265400, total 29526000+1265400=30791400. Then divide by 24: 30791400 ÷ 24 = 1282975. Then 178365 ÷ 1282975 ≈ 0.1390. Ah, so that's approximately 0.1390. So answer is D) 0.1390.
# Answer: D) 0.1390

### Question 19
# Explanation:
## Step1: Number of women and total subjects
Number of women: 456. Total subjects: 954.
## Step2: Probability first person is woman, then second person is woman (without replacement)
Probability first woman: $\frac{456}{954}$. Probability second woman (after first woman is selected): $\frac{455}{953}$. Multiply these probabilities: $\frac{456}{954} 	imes \frac{455}{953}$.
Calculate: $\frac{456 	imes 455}{954 	imes 953} = \frac{207480}{910162} \approx 0.2280$? Wait, no: 456×455=456×(450+5)=456×450 + 456×5=205200 + 2280=207480. 954×953=954×(950+3)=954×950 + 954×3=906300 + 2862=909162. Then 207480 ÷ 909162 ≈ 0.2282? Wait, 207480 ÷ 909162: let's divide numerator and denominator by 6: 34580 ÷ 151527 ≈ 0.2282. So answer is A) 0.2282.
# Answer: A) 0.2282

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QUESTION IMAGE

Question

Explanation:

Question 13

Step1: Find probability of sum 7 in one roll

Step2: Probability of sum 7 three times (independent events)

Step1: Probability one witness picks a man

Step2: Probability all 4 witnesses pick the same man

Step1: Probability first card is King

Step2: Probability second card is Queen (without replacement)

Step3: Multiply the probabilities (independent? No, dependent, so use multiplication rule for dependent events: $P(K_1 \cap Q_2) = P(K_1) \times P(Q_2|K_1)$

Answer:

Question 14