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For two independent events, A and B, P (A) = .4 and P(B) = .2 :

a. Find P (A∩B)

b. Find P (A/B)

c. Find P (A∪B)

Short Answer

Expert verified

Answer

  1. 0.08
  2. 0.4
  3. 0.52

Step by step solution

01

Step-by-Step SolutionStep 1: Introduction

There are two occurrences, A and B. If the outcome of the A event has no impact on the outcome of the B, the two occurrences are said to be independent.

P(A∩B)= P(A)×P(B)

02

Find the required probability

Since A and B are independent events.

Therefore,

P (A∩B) = P(A)×P(B)= .4×.2= .08

Hence, the required probability is 0.08.

03

Find the required probability

P (A/B) =P (A∩B)P (B)=.08.2=.4

Hence, the required probability is 0.4.

04

Find the required probability

P (A∪B) = P(A) + P (B)−P(A∩B)= 0.4 + 0.2−0.08= 0.52

Hence, the required probability is 0.52.

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Most popular questions from this chapter

Jamming attacks on wireless networks. Refer to the International Journal of Production Economics (Vol. 172, 2016) study of U.S. military jamming attacks on wireless networks used by terrorists, Exercise 2.8 (p. 73). Recall that 80 recent jamming attacks were classified according to network type (WLAN, WSN, or AHN) attacked and the network's number of channels (single- or multi-channel). The results are reproduced in the accompanying table.

a. Find the probability that a recent jamming attack involved a single-channel network.

b. Find the probability that a recent jamming attack involved a WLAN network.


Network Type/Number of Channels

Number of Jamming Attacks

WLAN / Single

31

WSN / Single

13

AHN / Single

8

WLAN / Multi

14

WSN / Multi

9

AHN / Multi

5

TOTAL

80

Source: S. Vadlamani et al., "Jamming Attacks on Wireless Networks: A Taxonomic Survey, "International Journal of Production Economics, Vol. 172, 2016 (Figure 6)

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P (E1) = .20, P (E2) = .30, P (E3)= .30, P (E4) = .10, P (E5) = .10.

a. Calculate P (A), P (B), and P (A∩B).

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c. Calculate the conditional probabilityP (E1/A)in two ways: (1) Add the adjusted (conditional) probabilities of the sample points in the intersection A∩B, as these represent the event that B occurs given that A has occurred; (2) use the formula for conditional probability:

P (B/A) =P (A∩B)P (A)

Verify that the two methods yield the same result.

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