MGMT2262

Worksheet #2 Solutions

First, we’ll make up a crosstab:

	Bay	Not Bay	Total
Walmart	7	33	40
Not Walmart	8	52	60
Total	15	85	100

1) P(Bay or Walmart) = P(Bay) + P(Walmart) - P(Bay and Walmart) = 15% + 40% - 7% = 48%

2) P(Walmart | Bay) = P(Bay and Walmart)/P(Bay) = 7/15 = 46.67%

3) P(Walmart and not Bay) = 33% from the crosstab

4) P(not Bay | Walmart) = P(Walmart and not Bay)/P(Walmart) = 33/40 = 82.5%

5) P(not Walmart | not Bay) = P(not Bay and not Walmart)/P(not Bay) = 52/85 = 61.18%

6) P(neither Bay nor Walmart) = 52% from crosstab. The other solution is to use DeMorgan’s Law: P(neither Bay nor Walmart) = 100% - P(Bay or Walmart) = 100% - 48% = 52%.

7) P(Bay | Walmart) = 7/40 = 17.5%; P(Bay | not Walmart) = 8/60 = 13.33%. So, someone who shops at Walmart is more likely to shop at The Bay than someone who doesn’t shop at Walmart.

8) We have P(Bay and Walmart) = 7%. P(Bay)*P(Walmart) = (0.15)(0.4) = 0.06 = 6%. Since these are not equal, the events are not independent and so shopping at one store depends on shopping at the other. One other solution is to note that P(Bay | Walmart) = 17.5% from question 7. However, P(Bay) = 15%. Since P(Bay) doesn’t equal P(Bay | Walmart), we reach the same conclusion.

9) Since P(Bay and Walmart) = 7% is not zero, the events of shopping at the 2 stores are not mutually exclusive.

10) P(revenue ³ 500K) = (438 + 365)/1000 = 803/1000 = 80.3%

11) P(profit ³ 100K | revenue ³ 500K) = (81 + 98 + 11 + 67)/(438 + 365) = 257/803 = 32%

12) P(profit < 100K) = (268 + 473)/1000 = 741/1000 = 74.1%

13) P(revenue < 500K | profit < 100K) = (141 + 54)/(268 + 473) = 195/741 = 26.32%

14) Revenue under $500,000 and net profit of $500,000 or more are mutually exclusive since the crosstab value is zero. These categories must be mutually exclusive since you can’t have a net profit of $500,000 or more if your revenue is less than that.

15) P(revenue ³ 500K or profit ³ 100K) = P(revenue ³ 500K) + P(profit ³ 100K) – P(revenue ³ 500K and profit ³ 100K) = (438 + 365)/1000 + (181 + 78)/1000 – (81 + 98 + 11 + 67)/1000 = 805/1000 = 80.5%

16) This table summarizes the results:

Profit|Revenue	< $500K	$500K - $1000K	$1000K +
$100K - $500K	2	81	98
$500K +	0	11	67
Total	197	438	365
	1.02%	21.00%	45.21%

For example, for revenue under $500,000, the percentage of these companies with net profit of $100,000 or more is (2 + 0)/197 = 1.02%. As revenues increase, the percentage of companies with net profit of $100,000 or more increases.

17) P(revenue ³ 500K and profit ³ 100K) = (81 + 98 + 11 + 67)/1000 = 0.257

P(revenue ³ 500K)*P(profit ³ 100K) = [(438 + 365)/1000]* [(181 + 78)/1000] = (0.803)(0.259) = 0.208. Since the two sides of the equation are not equal, having profit of $100,000 or more depends on having revenue of $500,000 or more.

18) We first need to compute the overall market share for the under 25 age group. This Excel output summarizes the results:

Brand	% market share	% used by under 25 age group	Weighted average
Brand A	0.25	0.7	0.175
Brand B	0.15	0.55	0.0825
Brand C	0.6	0.4	0.24
		Total	0.4975

We see the overall market share for the under 25 age group is 49.75%. We see that Brand C comprises the largest percentage at 24%. Therefore, the percentage of the under 25 age group that uses Brand C is 24/49.75 = 48.24%.

19) 12P4 = 11,880. The other method is 12x11x10x9 = 11,880

20) 12C3 = 220

21) 36P4 = 1,413,720. The other method is 36x35x34x33 = 1,413,720

22) 26x35x34x33 = 1,021,020. The other method is 26 x 35P3 = 26 x 39,270 = 1,021,020.

From the textbook

3.11 (page 99)

Part a

P(M) = 0.25, P(V) = 0.4, P(M and V) = 0.1

Part b

	M	Not M	Total
V	0.1	0.3	0.4
Not V	0.15	0.45	0.6
Total	0.25	0.75	1.0

Part c

P(M or V) = 0.25 + 0.4 – 0.1 = 0.55

P(not M and not V) = 0.45

P(exactly one) = 0.15 + 0.3 = 0.45

3.24 (page 106)

Part a

P(M) = 0.15, P(MBA) = 0.25, P(MBA | M) = 0.6

P(M and MBA) = P(M)*P(MBA | M) = (0.15)(0.6) = 0.09

Setting up the crosstab:

	M	Not M	Total
MBA	0.09	0.16	0.25
Not MBA	0.06	0.69	0.75
Total	0.15	0.85	1.0

P(M | MBA) = 0.09/0.25 = 0.36

Part b

Approach 1

LS = P(M and MBA) = 0.09

RS = P(M)*P(MBA) = (0.15)(0.25) = 0.0375

LS does not equal RS. Therefore, these are not independent events

Approach 2

LS = P(MBA | M) = 0.6

RS = P(MBA) = 0.25

Again, same result.

3.38 (page 111)

Part a

P(S) = 0.6 P(pass | S) = 0.85 P(S and pass) = (0.6)(0.85) = 0.51

P(not S) = 0.4 P(pass | not S) = 0.1 P(not S and pass) = (0.4)(0.1) = 0.04

P(pass) = 0.51 + 0.04 = 0.55

P(S | pass test) = 0.51/0.55 = 0.9273

Part b

Based on the results from part a, the test appears to be, by and large, a valuable way to screen applicants.

I also looked at the problem from the perspective of seeing the percentage who succeed given that they fail the test:

P(S) = 0.6 P(fail | S) = 0.15 P(S and fail) = (0.6)(0.15) = 0.09

P(not S) = 0.4 P(fail | not S) = 0.9 P(not S and fail) = (0.4)(0.9) = 0.36

P(fail) = 0.09 + 0.36 = 0.45

P(S | fail) = 0.09/0.45 = 0.2

Of those who fail the test, only 20% are successful managers. Again, based on this perspective, the test appears to be a valuable way to screen applicants.