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vova2212 [387]
3 years ago
10

The sample space of a random experiment is {a, b, c, d, e} with probabilities 0.1, 0.1, 0.2, 0.4, and 0.2 respectively. Let A de

note the event {a, b, c}, and let B denote the event {c, d, e}. Determine:
a. P(A)b. P(B)c. P(A’)d. P(AUB)e. P(AnB)
Mathematics
1 answer:
antoniya [11.8K]3 years ago
8 0

Answer:

a) P(A) =P(a)+P(b) +P(c)= 0.1+0.1+0.2 = 0.4

b) P(B) =P(c) +P(d)+P(e)=0.2+0.4+0.2=0.8

c) P(A') = 1-P(A) =1-0.4=0.6

d) P(A \cup B) =0.4 +0.8-0.2 =1.0

e)  The intersection between the set A and B is the element c so then we have this:

P(A \cap B) = P(c) =0.2

Step-by-step explanation:

We have the following space provided:

S= [a,b,c,d,e]

With the following probabilities:

P(a) =0.1, P(b)=0.1, P(c) =0.2, P(d)=0.4, P(e)=0.2

And we define the following events:

A= [a,b,c], B=[c,d,e]

For this case we can find the individual probabilities for A and B like this:

P(A) = 0.1+0.1+0.2 = 0.4

P(B) =0.2+0.4+0.2=0.8

Determine:

a. P(A)

P(A) =P(a)+P(b) +P(c)= 0.1+0.1+0.2 = 0.4

b. P(B)

P(B) =P(c) +P(d)+P(e)=0.2+0.4+0.2=0.8

c. P(A’)

From definition of complement we have this:

P(A') = 1-P(A) =1-0.4=0.6

d. P(AUB)

Using the total law of probability we got:

P(A \cup B) =P(A) +P(B)-P(A \cap B)

For this case P(A \cap B) = P(c) =0.2, so if we replace we got:

P(A \cup B) =0.4 +0.8-0.2 =1.0

e. P(AnB)

The intersection between the set A and B is the element c so then we have this:

P(A \cap B) = P(c) =0.2

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