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Would you please help me? am cofused

1 Answer

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SOLUTION

The probabilty of an event is the chances or likelyhood that the event will happen.

Consider the image below

From the image above, to obtain the probability of each event(tick mark) be label A, B, C , D, and E. we locate the position of each on the number line and divide by the total which is 10.

Hence,

The probability of tick mark A is

Since event A is on 0


\text{Pr(A)}=(0)/(10)=0

Pr(A)=0 (impossible)

For Probability of tick mark B,

B is on the 5th line, hence, the probability will be


pr(B)=(5)/(10)=(1)/(2)=0.5

Pr(B) = 1/2 ( Neither unlikely or likely)

For the probability of tick mark C,

C is on the 10th line, hence the probability will be


pr(C)=(10)/(10)=1

Pr(C) = 1 (certain)

For the probability of tick mark D.we have D is located on 2

Hence, the probability will be


Pr(D)=(2)/(10)=(1)/(5)=0.2

Pr(D) = 1/5 (unlikely)

For the probability of tick mark E.

E is located on 9th hence th probability will be


Pr(E)=(9)/(10)=0.9

Pr(E) = (9/10) Likely

Would you please help me? am cofused-example-1
User Kiran Malvi
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