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A sample space consists of 9 elementary outcomes e1, e2, ..., e9 whose probabilities are

P(e1) = P(e2) = .08, P(e3) = P(e4) = P(e5) = .1
P(e6) = P(e7) = .2, P(e8) = P(e9) = .07
Suppose A = {e1, e5, e8}, B = {e2, e5, e8, e9}
(a) Calculate P (A), P (B), and P (A ∩ B)
(b) Using the addition law of probability, calculate P (A ∪ B)
(c) List the composition of the event A ∪ B, and calculate P (A ∪ B) by adding the probabilities of the elementary outcomes.
(d) Calculate P (\bar{B}) from P (B), also calculate P (\bar{B}) directly from the elementary outcomes of \bar{B}

Answers (1)

Given data:

S = {e1, e2, e3, e4, e5, e6, e7, e8, e9}

A = {e1, e5, e8}

B = { e2, e5 e8, e9}

P (e1) = P (e2) = 0.08

P (e3) = P (e4) = P (e5) = 0.1

P (e6) = P (e7) = 0.2

P (e8) = P (e9) = 0.07

  1.  P (A), P (B) & P (A ∩ B)

A = {e1, e5, e8}  …….. (given)

Thus,

P (A) = P (e1) + P (e5) + P (e8)

         = 0.08 + 0.1 + 0.07

                  = 0.25

Now, B = { e2, e5 e8, e9}    ……. (given)

P (B) = P (e2) + P (e5) + P (e8) + P (e9)

Now, P (A ∩ B)

A ∩ B = {e5, e8}

Thus, P (A ∩ B) = P (e5) + P (e8)

                        = 0.1 + 0.07

                        = 0.17

  1. P (A U B)

P (A) = 0.25

P (B) = 0.32

P (A ∩ B) = 0.17

Now, P (A U B) = P (A) + P (B) – P (A ∩ B)        ……. (by general addition rule)

= 0.25 + 0.32 – 0.17

= 0.40

  (c) A = {e1, e5, e8}

       B = { e2, e5 e8, e9}        ……. (given)

Thus, A U B = {e1, e2, e5, e8, e9}

Thus, P (A U B) = P (e1) + P (e2) + P (e5) + P (e8) + P (e9)

                        = 0.08 + 0.08 + 0.1 + 0.07 + 0.07

                        = 0.40

(d) P (\bar{B})

P (\bar{B}) = 1 – 0.32

            = 0.68

Now, we have,

B = { e2, e5 e8, e9}       

Thus, \bar{B} = { e1, e3, e4, e6, e7}

P (\bar{B}) = P (e1) + P (e3) + P (e4) + P (e6) + P (e7)

            = 0.08 + 0.1 + 0.1 + 0.2 +0.2

            = 0.68

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