Problems 9-15 please! s rofiled, what is the bals that are numbered a team of S

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Problems 9-15 please! s rofiled, what is the bals that are numbered a team of S sbudents wwili be selected. o has three stages. Stage I can he to 6 months. Stage 11l can be inished in 3 to9 months. State III can be inished in 4 to monthswmany ounconmes are therer 6. What is the probabálity that a card drawn n d digits (There are 5 2 cards inom ades k of cards will be either a face card or a Diamond? Put answer deck of cards, there are 13 cards in each suite, face cards are Jack, Queen, King) poll asked 20,000 respondents to 7. A preseason sond a esm0 al hnne chapionshije oer the followisg question. will the SEC or the ACC conference 000 said the SEC would, 9000 said the ACC would, and 4000 said that neither said that either the SEC or the ACC will sen ould end a team. What is the 8. There is a 40 percent ehance that a certain s are parchased at the same time, what is the probability that erad of tire will last 40,000 miles, If four tires are t least one tire will need to be replaced before 40,000 (put answer ind digits) 9A basketball player is a 60% free throw What is the probability that she will make 1 point when she takes 2 free throws? 10. A and B are independent events. The 11. IE A and B are independent events with P(A).s and P(B)-0.6, then PAB)- P(A)- 65 and the PCA and B) 26. What is P(B)? 12. If A and B are independent events with PA)-0.1S and P(B)-0.65, then P(A B)- 13. If A and B are mutually exclusive events with P(A)-0.2 and P(B) -0.5, then PAUB)- Answer the next two questions with the followin lity table 34 06 04 .09 .02 Rh 06 14. What is the probability that a married couple with both have type B blood (answer in 4 digits)? 1s. What is the probability that a person will be O on the condition that it is known that he or she has type Rh+ blood (answer in 4 digits)?

Explanation / Answer

9)

P(X=1) =P(first throw success and second failure+second throw success and first failure) =0.6*0.4+0.4*0.6 =0.48

10)

P(B) =P(A and B)/P(A) =0.26/0.65 =0.4

11)

P(A n B) =P(A)*P(B) =0.5*0.6 =0.30

12)

P(A|B) =P(A) =0.15

13)

P(A U B) =P(A)+P(B) =0.2+0.5 =0.7

14)probability of type B =0.09+0.02 =0.11

probability that noth have type B blood group =P(first have type B and second have type B) =0.11*0.11 =0.0121

15)

P(O|Rh+) =0.38/(0.34+0.09+0.04+0.38)=0.4471

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