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Activity: What are the Chances? Part E Finding Probabilities t. Decide an the pr

ID: 2910000 • Letter: A

Question

Activity: What are the Chances? Part E Finding Probabilities t. Decide an the probability for each event below, Give your answer as a decimal value from .0to 1.0. (Some of the probabilities will be approximate ) 2. Describe how you decided on the probability for the events, including whether your answer was based on a theoretical model (sample space! oron observed results based on experience or relative frequency A. Reaching into a bag with three red gum balls, two bue gum balls, and four black gum balls, and pulling out a blue gum bail B. Snow faling sometime in July in Florida C Snow falling sometime in July in New Zealand D. Flipping a coin twice and getting a head and a tail. E. Rolling a die and getting an odd number F. Going to a store and finding they don't have shoes in your sice G. Rolling two dice and getting doubles (same number on both dice). H. Rolling a pair of dice and getting doubles by the third roll Ian rall 1 or rall 2 or roll 3) Part II: Probabilities on the Number Line Make a nu place on the number line to indicate its probability mber line like the one below and put the letter of each of the events above in the proper

Explanation / Answer

A. bag contains 3 red gum balls,2 blue and 4 black .

total no. of balls in the bag=3+2+4=9.

so, p( getting a blue gum ball)=(no. of blue gum balls)/( total no. of balls)

=(2÷9)=0.22(approx.)

B. snow falling is a subjective probability .no definite rule can be applied for it .so the probability can not be determined.

C. same argument as the above.

D. sample space={(H,H),(H,T),(T,H),(T,T)}

total no. of points in sample space =4

no. of favorafavo points =2 {( H,T) and(T,H)}

hence probability=2/4=0.5(ans)

E. no. of points in sample space=6. {1,2,34,5,6}

no. of even points =3{2,4,6}

so the probability=3/6=0.5(ans)

G. total no. pf points in sample space=36

no. of favourable points{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}

i.e 6 points

hence required probability=6/36=0.16(ans)

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