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Reptile [31]
3 years ago
12

In an experiment, A,B, C, andD are events with probabilitiesP[A UB] = 5/8,P[A] =3/8,

Mathematics
1 answer:
BARSIC [14]3 years ago
7 0

Answer:

Step-by-step explanation:

Hello!

You have 4 events A, B, C and D

With probabilities:

P(A∪B)= 5/8

P(A)= 3/8

P(C∩D)= 1/3

P(C)= 1/2

A and B are disjoint events, this means that there are no shared elements between then and their intersection is void, symbolically A∩B= ∅, in consequence, these events are mutually exclusive.

C and D are independent events, this means that the occurrence of one of them does not affect the probability of occurrence of the other one in two consecutive repetitions.

a.

i. P(A∩B)= 0

⇒ Since A and B are disjoint events, the probability of their intersection is zero.

ii. A and B are mutually exclusive events, this means that P(A∪B)= P(A)+P(B)

⇒ From this expression, you can clear the probability of b as P(B)= P(A∪B)-P(A)= 5/8-3/8= 1/4

iii. If Bc is the complementary event of B, its probability would be P(Bc)= 1 - P(B)= 1 - 1/4= 3/4. If the events A and B are mutually exclusive and disjoint, it is logical to believe that so will be the events A and Bc, so their intersection will also be void:

P(A∩Bc)= 0

vi.P(A∪Bc)= P(A) + P(Bc)= 3/8+3/4= 9/8

b.

If A and B are independent then the probability of A is equal to the probability of A given B, symbolically:

P(A)= P(A/B)

P(A/B)= \frac{P(AnB)}{P(B)}= \frac{0}{1/4}= 0

P(A)= 3/8

P(A) ≠ P(A/B) ⇒ A and B are not independent.

c.

i. P(D) ⇒ Considering C and D are two independent events, then we know that P(C∩D)= P(C)*P(D)

Then you can clear the probability of D as:

P(D)= P(C∩D)/P(C)= (1/3)/(1/2)= 2/3

ii. If Dc is the complementary event of D, then its probability is P(Dc)= 1 - P(D) = 1 - 2/3= 1/3

P(C∩Dc)= P(C)*P(Dc)= (1/2)*(1/3)= 1/6

iii. Now Cc is the complementary event of C, its probability is P(Cc)= 1 - P(C)= 1 - 1/2= 1/2

P(Cc∩Dc)= P(Cc)*P(Dc)= (1/2)*(1/3)= 1/6

vi. and e.

P(C/D)= \frac{P(CnD)}{P(D)} = \frac{1/3}{2/3} = 1/2

P(C)=1/2

As you can see the P(C)=P(C/D) ⇒ This fact proves that the events C and D are independent.

d.

i. P(C∪D)= P(C) + P(D) - P(C∩D)= 1/2 + 2/3 - 1/3= 5/6

ii. P(C∪Dc)= P(C) + P(Dc) - P(C∩Dc)= 1/2 + 1/3 - 1/6= 2/3

I hope it helps!

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Answer:

For 1948 Men's :Interquartile range is 1.5, Median is 58.3

For 2012 Men's: Interquartile range is 0.315, Median is 47.86

You can infer that in 2012 the swimmers were better and it was more competitive as the interquartile range was lower and the median was also lower as well

Step-by-step explanation:

1948 Men's 100m

57.3, 57.8, 58.1, 58.3, 58.3, 59.3, 59.6,1:00.5

               ⬆                ⬆                 ⬆

         Low IQR      Median       High IQR

Low IQR is average of 57.8 and 58.1 = 57.95

High IQR  is average of 59.3 and 59.6 =  59.45

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Interquartile range is High IQR- Low IQR

59.45 - 57.95=1.5

2012 Men's 100m

47.52, 47.53, 47.8, 47.84, 47.88, 47.92, 48.04, 48.44

                    ⬆                  ⬆                      ⬆

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Low IQR is average of 47.53 and 47.8 = 47.665

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Read more on Brainly.com - brainly.com/question/14915771#readmore

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Answer:

Final answer is x-6y=23

Step-by-step explanation:

We need to find the equation of the line that is parallel to x=6y-5 and that passes through (5,-3).

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