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Chapter 1: Q11SE (page 54)

Let\({{\bf{A}}_{\bf{1}}}\),\({{\bf{A}}_{\bf{2}}}\), and\({{\bf{A}}_{\bf{3}}}\) be three arbitrary events. Show that the probability that exactly one of these three events will occur is

\({\bf{Pr}}\left( {{{\bf{A}}_{\bf{1}}}} \right){\bf{ + Pr}}\left( {{{\bf{A}}_{\bf{2}}}} \right){\bf{ + Pr}}\left( {{{\bf{A}}_{\bf{3}}}} \right){\bf{ - 2Pr}}\left( {{{\bf{A}}_{\bf{1}}} \cap {{\bf{A}}_{\bf{2}}}} \right){\bf{ - 2Pr}}\left( {{{\bf{A}}_{\bf{1}}} \cap {{\bf{A}}_{\bf{3}}}} \right){\bf{ - 2Pr}}\left( {{{\bf{A}}_{\bf{2}}} \cap {{\bf{A}}_{\bf{3}}}} \right){\bf{ + 3Pr}}\left( {{{\bf{A}}_{\bf{1}}} \cap {{\bf{A}}_{\bf{2}}} \cap {{\bf{A}}_{\bf{3}}}} \right)\)

Short Answer

Expert verified

\(\begin{aligned}P &\left( {{A^C}_1 \cap {A^C}_2 \cap {A_3}} \right) \cup P\left( {{A^C}_1 \cap {A^C}_2 \cap {A_3}} \right) \cup P\left( {{A^C}_1 \cap {A^C}_2 \cap {A_3}} \right)\\ &= \left( {P\left( {{A_1}} \right) + P\left( {{A_2}} \right) + P\left( {{A_3}} \right) - 2P\left( {{A_1} \cap {A_2}} \right) - 2P\left( {{A_1} \cap {A_3}} \right) - 2P\left( {{A_2} \cap {A_3}} \right) + 3P\left( {{A_1} \cap {A_2} \cap {A_3}} \right)} \right)\end{aligned}\)

Step by step solution

01

Given information

\({A_1},{A_2},{A_3}\) are three arbitrary events.

02

Calculate the probability\({\bf{P}}\left( {{{\bf{A}}_{\bf{1}}} \cap {{\bf{A}}^{\bf{C}}}_{\bf{2}} \cap {{\bf{A}}^{\bf{C}}}_{\bf{3}}} \right)\)

\(\begin{aligned}P\left( {{A_1} \cap {A^C}_2 \cap {A^C}_3} \right) &= P\left( {{A_1}} \right) \cap P\left( {{A^C}_2} \right) \cap P\left( {{A^C}_3} \right)\\ &= P\left( {{A_1}} \right)P\left( {{A^C}_2} \right)P\left( {{A^C}_3} \right)\\ &= P\left( {{A_1}} \right)\left( {1 - P\left( {{A_2}} \right)} \right)\left( {1 - P\left( {{A_3}} \right)} \right)\\ &= P\left( {{A_1}} \right)\left( {1 - P\left( {{A_2}} \right) - P\left( {{A_3}} \right) + P\left( {{A_2}} \right)P\left( {{A_3}} \right)} \right)\\ &= P\left( {{A_1}} \right) - P\left( {{A_1}} \right)P\left( {{A_2}} \right) - P\left( {{A_1}} \right)P\left( {{A_3}} \right) + P\left( {{A_1}} \right)P\left( {{A_2}} \right)P\left( {{A_3}} \right)\end{aligned}\)

03

Calculate the probability\({\bf{P}}\left( {{{\bf{A}}^{\bf{C}}}_{\bf{1}} \cap {{\bf{A}}_{\bf{2}}} \cap {{\bf{A}}^{\bf{C}}}_{\bf{3}}} \right)\)

\(\begin{aligned}P\left( {{A^C}_1 \cap {A_2} \cap {A^C}_3} \right) &= P\left( {{A^C}_1} \right) \cap P\left( {{A_2}} \right) \cap P\left( {{A^C}_3} \right)\\ &= P\left( {{A^C}_1} \right)P\left( {{A_2}} \right)P\left( {{A^C}_3} \right)\\ &= \left( {1 - P\left( {{A_1}} \right)} \right)P\left( {{A_2}} \right)\left( {1 - P\left( {{A_3}} \right)} \right)\\ &= \left( {1 - P\left( {{A_1}} \right)} \right)\left( {P\left( {{A_2}} \right) - P\left( {{A_2}} \right)P\left( {{A_3}} \right)} \right)\\ &= P\left( {{A_2}} \right) - P\left( {{A_1}} \right)P\left( {{A_2}} \right) - P\left( {{A_2}} \right)P\left( {{A_3}} \right) + P\left( {{A_1}} \right)P\left( {{A_2}} \right)P\left( {{A_3}} \right)\end{aligned}\)

04

Calculate the probability\({\bf{P}}\left( {{{\bf{A}}^{\bf{C}}}_{\bf{1}} \cap {{\bf{A}}^{\bf{C}}}_{\bf{2}} \cap {{\bf{A}}_{\bf{3}}}} \right)\)

\(\begin{aligned}P\left( {{A^C}_1 \cap {A^C}_2 \cap {A_3}} \right) &= P\left( {{A^C}_1} \right) \cap P\left( {{A^C}_2} \right) \cap P\left( {{A_3}} \right)\\ &= P\left( {{A^C}_1} \right)P\left( {{A^C}_2} \right)P\left( {{A_3}} \right)\\ &= \left( {1 - P\left( {{A_1}} \right)} \right)\left( {1 - P\left( {{A_2}} \right)} \right)P\left( {{A_3}} \right)\\ &= \left( {1 - P\left( {{A_1}} \right)} \right)\left( {P\left( {{A_3}} \right) - P\left( {{A_2}} \right)P\left( {{A_3}} \right)} \right)\\ &= P\left( {{A_3}} \right) - P\left( {{A_1}} \right)P\left( {{A_3}} \right) - P\left( {{A_2}} \right)P\left( {{A_3}} \right) + P\left( {{A_1}} \right)P\left( {{A_2}} \right)P\left( {{A_3}} \right)\end{aligned}\)

05

Calculate the probability

\(\begin{aligned}P &\left( {{A^C}_1 \cap {A^C}_2 \cap {A_3}} \right) \cup P\left( {{A^C}_1 \cap {A^C}_2 \cap {A_3}} \right) \cup P\left( {{A^C}_1 \cap {A^C}_2 \cap {A_3}} \right)\\ &= P\left( {{A^C}_1 \cap {A^C}_2 \cap {A_3}} \right) + P\left( {{A^C}_1 \cap {A^C}_2 \cap {A_3}} \right) + P\left( {{A^C}_1 \cap {A^C}_2 \cap {A_3}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{\left( {P\left( {{A_1}} \right) - P\left( {{A_1}} \right)P\left( {{A_2}} \right) - P\left( {{A_1}} \right)P\left( {{A_3}} \right) + P\left( {{A_1}} \right)P\left( {{A_2}} \right)P\left( {{A_3}} \right)} \right) + }\\{\left( {P\left( {{A_2}} \right) - P\left( {{A_1}} \right)P\left( {{A_2}} \right) - P\left( {{A_2}} \right)P\left( {{A_3}} \right) + P\left( {{A_1}} \right)P\left( {{A_2}} \right)P\left( {{A_3}} \right)} \right) + }\\{\left( {P\left( {{A_3}} \right) - P\left( {{A_1}} \right)P\left( {{A_3}} \right) - P\left( {{A_2}} \right)P\left( {{A_3}} \right) + P\left( {{A_1}} \right)P\left( {{A_2}} \right)P\left( {{A_3}} \right)} \right)}\end{array}} \right)\\ &= \left( {P\left( {{A_1}} \right) + P\left( {{A_2}} \right) + P\left( {{A_3}} \right) - 2P\left( {{A_1}} \right)P\left( {{A_2}} \right) - 2P\left( {{A_1}} \right)P\left( {{A_3}} \right) - 2P\left( {{A_2}} \right)P\left( {{A_3}} \right) + 3P\left( {{A_1}} \right)P\left( {{A_2}} \right)P\left( {{A_3}} \right)} \right)\end{aligned}\)

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Most popular questions from this chapter

Use the data on dishwasher shipment in Table \({\bf{11}}.{\bf{13}}\;\)on page \({\bf{744}}\) .suppose that we wish to fit a multiple linear regression model for predicting dishwasher shipment from time (the year 1960) and private residential investment. Suppose that the parameters have the improper prior proportional to \({\raise0.7ex\hbox{\(1\)} \!\mathord{\left/ {\vphantom {1 \tau }}\right.\\} \!\lower0.7ex\hbox{\(\tau \)}}\) the use of the Gibbs sampling algorithm to obtain a sample of size 10,000 from the joint posterior distribution of the parameters.

a. Let \({\beta _{1\,}}\) be the coefficient of time. Draw a plot of sample c.d.f of \(\left| {{\beta _{1\,}}} \right|\) using your posterior sample.

b. We are interested in the values of your posterior distribution 1986.

i. Draw a histogram of the values \({{\bf{\beta }}_{{\bf{0}}\,}}{\bf{ + 26}}{{\bf{\beta }}_{{\bf{1}}\,}}{\bf{ + 67}}{\bf{.2}}{{\bf{\beta }}_{{\bf{2}}\,}}\) from your posterior distribution.

ii. For each of your simulated parameters, simulate a dishwasher sales figure for

1986 (time = 26 and private residential investment = 67.2). compute a 90 percent prediction interval from the simulated values and compare it to the interval found in Example 11.5.7.

iii. Draw a histogram of the simulated 1986 sales figures, and compare it to the histogram in part I. Can you explain why one sample seems to have a more considerable variance than the other?

For the conditions of Exercise 4, what is the probability that neither student A nor student B will fail the examination?

Prove that for every two events A and B, the probability that exactly one of the two events will occur is given by the expression

\(\Pr \left( A \right) + \Pr \left( A \right) - 2\Pr \left( {A \cap B} \right)\)

If two balanced dice are rolled, what is the probability that the difference between the two numbers that appear will be less than 3?

Suppose that X has the binomial distribution with parameters n and p, and that Y has the negative binomial distribution with parameters r and p, where r is a positive integer. Show that \({\bf{Pr}}\left( {{\bf{X < r}}} \right){\bf{ = Pr}}\left( {{\bf{Y > n - r}}} \right)\)by showing

that both the left side and the right side of this equation can be regarded as the probability of the same event in a sequence of Bernoulli trials with probability p of success.
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