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In a sales effectiveness seminar, a group of salesrepresentatives tried two approaches to selling a customer a newautomobile: the aggressive approach and the passiv...

Question

In a sales effectiveness seminar, a group of salesrepresentatives tried two approaches to selling a customer a newautomobile: the aggressive approach and the passive approach. For1160 customers, the following record was kept: Sale No Sale Row Total Aggressive 256324580Passive 480100580Column Total 7364241160Suppose a customer is selected at random from the 1160participating customers. Let us use the following notation forevents: A = aggressiveapproach, Pa = passiveapproach, S = sale, N = nosale.

In a sales effectiveness seminar, a group of sales representatives tried two approaches to selling a customer a new automobile: the aggressive approach and the passive approach. For 1160 customers, the following record was kept: Sale No Sale Row Total Aggressive 256 324 580 Passive 480 100 580 Column Total 736 424 1160 Suppose a customer is selected at random from the 1160 participating customers. Let us use the following notation for events: A = aggressive approach, Pa = passive approach, S = sale, N = no sale. So, P(A) is the probability that an aggressive approach was used, and so on. (a) Compute P(S), P(S | A), and P(S | Pa). (Enter your answers as fractions.) P(S) = P(S | A) = P(S | Pa) = (b) Are the events S = sale and Pa = passive approach independent? Explain. No. The two events cannot occur together. No. P(S) ≠ P(S | Pa). Yes. P(S) = P(S | Pa). Yes. The two events can occur together. (c) Compute P(A and S) and P(Pa and S). (Enter your answers as fractions.) P(A and S) = P(Pa and S) = (d) Compute P(N) and P(N | A). (Enter your answers as fractions.) P(N) = P(N | A) = (e) Are the events N = no sale and A aggressive approach independent? Explain. Yes. The two events can occur together. Yes. P(N) = P(N | A). No. The two events cannot occur together. No. P(N) ≠ P(N | A). (f) Compute P(A or S). (Enter your answer as a fraction.) P(A or S) =



Answers

Expand Your Knowledge: Conditional Probability In the western United States, there are many dry-land wheat farms that depend on winter snow and spring rain to produce good crops. About $65 \%$ of the years, there is enough moisture to produce a good wheat crop, depending on the region (Reference: Agricultural Statistics, U.S. Department of Agriculture). (a) Let $r$ be a random variable that represents the number of good wheat crops in $n=8$ years. Suppose the Zimmer farm has reason to believe that at least 4 out of 8 years will be good. However, they need at least 6 good years out of 8 to survive financially. Compute the probability that the Zimmers will get at least 6 good years out of 8 , given what they believe is true; that is, compute $P(6 \leq r \mid 4 \leq r)$. See part (d) for a hint. (b) Let $r$ be a random variable that represents the number of good wheat crops in $n=10$ years. Suppose the Montoya farm has reason to believe that at least 6 out of 10 years will be good. However, they need at least 8 good years out of 10 to survive financially. Compute the probability that the Montoyas will get at least 8 good years out of 10 , given what they believe is true; that is, compute $P(8 \leq r \mid 6 \leq r)$. (c) List at least three other areas besides agriculture to which you think conditional binomial probabilities can be applied. (d) Hint for solution: Review item 6, conditional probability, in the summary of basic probability rules at the end of Section 4.2. Note that $$ P(A \mid B)=\frac{P(\text { Aand } B)}{P(B)} $$ and show that in part (a), $$ P(6 \leq r \mid 4 \leq r)=\frac{P((6 \leq r) \text { and }(4 \leq r))}{P(4 \leq r)}=\frac{P(6 \leq r)}{P(4 \leq r)} $$

Alright, So 29 is, um a conditional probability problem. And so conditional probabilities can be where you're given certain information and you Onley do your bottom number out of the totals from a column or a row. Now, part of this question does involve a total probability where you look at this grand total number, so we will address that. But this is talking about aggressive versus passive sales approaches and whether or not someone's gonna make a sale, uh, to a customer or not s Oh, this is just a random survey of that happening. So the first task were asked to dio is to calculate the probability of a sale occurring So the probability of a sale occurring total would include that 6 86 out of 11 60 so six that equal six 86 out of 11 60 or then asked to calculate the probability of getting a sale given that's what that bar means, that someone was aggressive. So given their aggressive, whatever I'm given, that's gonna be the row or column that I focus on. So we're gonna focus on the aggressive column and that's gonna be my top in my bottom. So making a sale is to 70 out of 5 80. Then the other one says the probability of making a sale, given that they are passive. So now that we focus on passive now, it's going to be 4 16 out of 5. 80. Okay? And so are the events of getting a sale and passive, um, approach independent. And the answer is no. And that's because the probability of getting a sale and the probability of getting a sale, given that they're passive, are not congruent so it's not independent. So the answer Toby is No. Now we'll go onto part C. S O. We know they're not congruent probabilities. Uh, see, we're going to compute the probability of them being aggressive and getting a sale. So that's when we focus on this to 70 out of the 11 60. So to 70 out of 11 60 and then the probability of being passive and getting a sale. So the probability of being passive and doing a sale is going to be this 4 16 out of 11 60. And so then we were asked to compute. Okay, the probability of n occurring, so of not a sale happening so no sale is for 74 out of 11. 60. Okay. Forcing four out of 11. 60. Probability of not getting a sale, given that they were aggressive. So we're going to focus again on this a row and no sale will be 3 10 out of 5. 80. Okay. And then the events, Uh, no sale and aggressive independent. The answer is no, because for 74 out of 11, 60 is not the same as the decimal of 3, 10 out of 5. 80. So that percentage is different, or that probability is different, so they can't be independent. So the answer to eat is no. And last but not least, we're going to calculate the probability of A or s occurring. So the probability of A or S is going to be where you have the probability of a occurring plus the probability of a sale occurring minus the overlap probability of A and s. So we have probability of a we already had from probably of a is, um to public give a occurring is 5 80 out of 11 60 Probability of s occurring is 6 86 a sale coring out of 11 60 then that overlap of A and S is to 70 out of 11. 6 week calculated that earlier. So to 70 out of 11. 60. So you're gonna add up. Your numerator is the 5. 80 the 6 86 and subtract 2 70. You should get 996 over 11. 60 as your probability. Thank you.

They were looking at, uh, at least one of them did not make any sales. So we're counting all the zeros so we can't across the row and and down the column. So we have a total of eight out of 20 which gives this point for now in part B. They want exactly three, so it's either a 03 or one and two. So there's four of them like that out of 20 Percy, there's four of them out of 20 for being. There's five out of 20 for E. Yeah, they're getting into A and B, so they have to meet the criteria for a end be when she's on Lee two out of 20. So the only thing that matches that would be three comma zero b and C there's no those zero out of 20 g A or B. Very a work bee. So we have, um, eight out of 20 last four out of 20 minus A and B, which is two out of 20. So that gives us 10 out of 20 which equals 0.5 age B or C four out of 20 or at 20 plus. We're on 20 now, he ends, he was zero. So this just equals eight out of 20. Which is wait for I is a given be so a and B over B. That would be a chai if you out of Ford. So a A and B is too. Okay, those, uh, j be given the soapy Andy, which is one over d, which is five. Okay, we have c given be so C and B, which was 0/4 people's zero. Okay. Not giving compliments. So the probability, um, be given a compliment equals two out of 12 Rich's 0.167 and ability of C given a compliment, just three at it. Well, or 0.25 and a or B or C. So we just add them all up where you dig probability for A or B or signature from a veteran question, we actually have toe count them. So when we count them, we get 13 out of 20 and then oh, are they ended their and be intersect? Because we haven't A and B together. You know that A and B equals one and okay, Um, yes, he ends the equals zero. They do not intersect

Involved Number of 107 we need to evaluate eight C four one by three Raised to the power four into two by three Arrested Power four which is the probability of making four sales. So 84 is a factual by for factorial for pictorial this is one by three years to the power four into one by three years to the power four into tourist REpower four eight Victory 8765 for pictorial by four, 321 for pictorial This will be won by the bases are the same one by 31 by three is the power will be added up Tourist a part four This will get canceled out 32 six for two times. So now we have to use a calculator to find the probability. This is are five into 10 into 7 70 17 to two days to the powerful that is 1120 development three years about eight three days without eighties 6561 Approximately it will be 1120 Divide buyer 6561 That is zero point 17 away Approximately. So this will be their answer. Thank you

Yeah, I remember 108. In which we need to evaluate eight c 41 by tourists. The par four and one by tourists, PAL four, which is the probability of sale with anyone. Customer is run by two. So vitality that sales representative makes four cells. Okay, so we have to calculate this. So this is a pictorial by four pictorial for pictorial one by two arrested power eight, 8765 for victory by 4321 for pictorial even by to arrest about eight. Three to the six for 2008. So this will be 70. Divided by divide by tourists or eight. 2008 is 256. All this can be certified by 128 which is approximately good fun people is the 0.273 273 So this will be done. So thank you so much


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