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888-Can anybody help me to solve these seven questionsyour helpful answers will

ID: 3182823 • Letter: 8

Question

888-Can anybody help me to solve these seven questionsyour helpful answers will be greatly appreciated
7. Three numbers are randomly selected once from 0 to 9. Find the probabilities of the following two events Ai three numbers don't contain 0 and 5; A2 three numbers don't contain 0 or 5 8. 6 students live a dorm, find the probabilities of the following events: (1) The birthday of at least one student is in October; (2) Among 6 students, the birthdays of 4 students are exactly in October; (3) Among 6 students, the birthdays of 4 students are exactly in a same month. 9. A batch of products contain products A, B, and C. Suppose that products A, B, and C account for 60 %,30%, and 10 of total products, respectively. Now, a product that randomly withdrawn is not product C, find the probability that this product is product A. 10. A dorm is equipped with two warning systems. The working probabilities of systems 1 and 2 are 0.92 and 0.93, respectively. Under the condition that system 1 is down, the probability that system 2 is working is 0.85. Find the probability that (1) systems 1 and 2 are working: (2) system 2 is down and system 1 is working: (3) system 1 is working given that system 2 is down. 11. Let 0

Explanation / Answer

7)

number selected from 0 to 9.

n=10

selected number=r=3

total possible ways to pick any 3 numbers= 10C3

=(10!) /((10-4)!*4!)

total possible ways to pick any 3 numbers=s=120

a) A1=three numbers dont containe 0 and 5

sample space containing 0 and 5=(0,5,1),(0,5,2),(0,5,3),(0,5,4),(0,5,6),(0,5,7),(0,5,8),(0,5,9)

total=8

number of A1=120-8=112

P(A1)= n(A1)/s

=112/120

P(A1)=0.9333

Probability of A1=three numbers dont containe 0 and 5 is 0.9333 or 93.3%.

b)

A2=three numbers dont containe 0 or 5

sample space containing 0 and 5=112

sample space containg 0 or 5 =112/2=56

P(A2)= n(A2)/s

=56/120

P(A2)=0.4667

Probability of A2=three numbers dont containe 0 or 5 is 0.4667 or 46.67%.

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