EXCEL DATA p 0.09 x P(X <= x) P(X = x) 1 0.0900 0.0900 2 0.1719 0.0819 3 0.2464

Question

EXCEL DATA

p 0.09 x P(X <= x) P(X = x) 1 0.0900 0.0900 2 0.1719 0.0819 3 0.2464 0.0745 4 0.3143 0.0678 5 0.3760 0.0617 6 0.4321 0.0562 7 0.4832 0.0511 8 0.5297 0.0465 9 0.5721 0.0423 10 0.6106 0.0385 11 0.6456 0.0350 12 0.6775 0.0319 13 0.7065 0.0290 14 0.7330 0.0264 15 0.7570 0.0240 16 0.7789 0.0219 17 0.7988 0.0199 18 0.8169 0.0181 19 0.8334 0.0165 20 0.8484 0.0150 21 0.8620 0.0136 22 0.8744 0.0124 23 0.8857 0.0113 24 0.8960 0.0103 25 0.9054 0.0094 26 0.9139 0.0085 27 0.9216 0.0078 28 0.9287 0.0071 29 0.9351 0.0064 30 0.9409 0.0058 31 0.9463 0.0053 32 0.9511 0.0048 33 0.9555 0.0044 34 0.9595 0.0040 35 0.9631 0.0036 36 0.9665 0.0033 37 0.9695 0.0030 38 0.9722 0.0027 39 0.9747 0.0025 40 0.9770 0.0023 41 0.9791 0.0021 42 0.9810 0.0019 43 0.9827 0.0017 44 0.9842 0.0016 45 0.9856 0.0014 46 0.9869 0.0013 47 0.9881 0.0012 48 0.9892 0.0011 49 0.9902 0.0010 50 0.9910 0.0009 8. 0/3.5 points I Previous Answers My Notes Assume that 5% of people have Type AB blood. A hospital blood bank needs Type AB blood so the blood bank begins checking potential donors. Use this Excel file geometric probabilities (enable the macro after opening the file) or this Excel file geometric probabilities 2 to assist you with the geometric probability calculations required to answer the questions below Question 1. What is the probability that the first Type AB donor is the 6th person checked? Use 4 decimal places 035 Question 2. On average, how many donors must be checked until the first donor with Type AB blood is found? Use 2 decimal places. Question 3. What is the probability that there is a Type AB donor among the first 6 donors checked? Use 3 decimal places. Question 4. What is the probability that we won't find a Type AB donor before the 10 person, that is, what is the probability that the first Type AB donor is not among the first 9 donors? Use 3 decimal places.

Explanation / Answer

a)

P(first AB doner is 6th person checked) = 0.95*0.95*0.95*0.95*0.95*0.05 =  0.0387

b)

14

c)

P(an AB doner among first 6) = 1- no donor among first 6 = 1- (0.95^6) = 0.265

d)

P(first 9 donors are non AB doners) = 0.95^9 = 0.630

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