Q.6.1 Verify Equation 1.2.P(a1<X鈮�... [FREE SOLUTION]

91影视

A First Course in Probability

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 432 pages

ISBN: 9780321794772

Chapter 6: Q.6.1 (page 275)

Verify Equation $(1.2)$ . $P (a_{1} < X 鈮� a_{2}, b_{1} < Y 鈮� b_{2}) = F (a_{2}, b_{2}) + F (a_{1}, b_{1}) 鈭� F (a_{1}, b_{2}) 鈭� F (a_{2}, b_{1})$

Short Answer

Expert verified

Equation is proved.

$P (a_{1} < X 鈮� a_{2}, b_{1} < Y 鈮� b_{2}) = F (a_{2}, b_{2}) + F (a_{1}, b_{1}) 鈭� F (a_{1}, b_{2}) 鈭� F (a_{2}, b_{1})$

Step by step solution

Equation :

A declaration (represented by the symbol =) that the values of two mathematical expressions are equal.

Explanation :

Let us define the events,

$A : \{X 鈮� a_{1}\} B : \{X 鈮� a_{2}\} C : \{Y 鈮� b_{1}\} D : \{Y 鈮� b_{2}\}$

$P [a_{1} < X < a_{2} 鈭� b_{1} 鈮� Y 鈮� b_{2}] = P [(B - A) 鈭� (D - C)] = P [B 鈭� (D - C) - A 鈭� (D - C)] . . . . . . (1) [B y d i s t r i b u t i v e l a w]$

We know that if $E 鈯� F 鈬� E 鈭� F = E$ , then

$P (F - E) = P (\overset{炉}{E} 鈭� F) = P (F) - P (E 鈭� F) = P (F) - P (E) . . . . . . . (2)$

Obviously $A 鈯� B 鈬� [A 鈭� (D - C)] 鈯� [B 鈭� (D - C)]$

Hence by using $(2)$ , we get from $(1)$

$P [a_{1} < X 鈮� a_{2} 鈭� b_{1} < Y 鈮� b_{2}] = P [B 鈭� (D - C)] - P [A 鈭� (D - C)] = P [(B 鈭� D) - (B 鈭� C)] - P [(A 鈭� D) - (A 鈭� C)] = P (B 鈭� D) - (B 鈭� C) - P (A 鈭� D) + P (A 鈭� C) . . . . . . (3)$

We have localid="1647341297556" $P (B 鈭� D) = P [X 鈮� a_{2} 鈭� Y 鈮� b_{2}] = F (a_{2}, b_{2})$

Similarly localid="1647341325673" $P (B 鈭� C) = F (a_{2}, b_{1}) P (A 鈭� D) = F (a_{1}, b_{2})$

Substitute these values in $(3)$ we get

$P (a_{1} < X 鈮� a_{2}, b_{1} < Y 鈮� b_{2}) = F (a_{2}, b_{2}) + F (a_{1}, b_{1}) 鈭� F (a_{1}, b_{2}) 鈭� F (a_{2}, b_{1})$

Hence proved.

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91影视

Verify Equation $(1.2)$ . $P (a_{1} < X 鈮� a_{2}, b_{1} < Y 鈮� b_{2}) = F (a_{2}, b_{2}) + F (a_{1}, b_{1}) 鈭� F (a_{1}, b_{2}) 鈭� F (a_{2}, b_{1})$

Short Answer

Step by step solution

Equation :

Explanation :

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91影视

Verify Equation 1.2.P(a1<X鈮�a2,b1<Y鈮�b2)=F(a2,b2)+F(a1,b1)鈭�F(a1,b2)鈭�F(a2,b1)

Short Answer

Step by step solution

Equation :

Explanation :

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Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Probability and Statistics

Discrete Mathematics

Logic and Functions

Applied Mathematics

Pure Maths

Study anywhere. Anytime. Across all devices.

Verify Equation $(1.2)$ . $P (a_{1} < X 鈮� a_{2}, b_{1} < Y 鈮� b_{2}) = F (a_{2}, b_{2}) + F (a_{1}, b_{1}) 鈭� F (a_{1}, b_{2}) 鈭� F (a_{2}, b_{1})$