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30 May 2021, 12:34
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The probability that event M will not occur is 0.5 and the probability that event R will not occur is 0.5. If events M and R cannot both occur, which of the following is the probability that either event M or event R will occur?
P(M or R) = P(M) + P(R) - P(M and R)
P(M or R) = .5 + .5 - 0
P(M or R) = 1
I've changed the question to the probability instead being 0.5. Using the same solution method many have given here, I get a nonesense answer. I still do not see why the answer is not 11/25.
P(M or R) = P(M) + P(R) - P(M and R)
P(M or R) = .5 + .5 - 0
P(M or R) = 1
I've changed the question to the probability instead being 0.5. Using the same solution method many have given here, I get a nonesense answer. I still do not see why the answer is not 11/25.
11 Sep 2021, 09:17
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I got the the textbook solution. I don't get what's wrong with the alternative solution.
P(A or B) = 1- P(not A AND not B) = 1 - P(not A)* P(not B) = 1 - 4/5*3/5 = 13/25
What am I doing wrong? What's the difference with this problem, in which it works:
What is the probability of NOT getting 1 in at least one two independent dice rolls:
Textbook solution: P= P(1) + P(1) + P (1 and 1) = 1/6 + 1/6 - 1/36 = 11/36
Alternative: P= 1 - P(not 1)* P(not 1)= 1- 5/6 * 5/6 = 11/36
P(A or B) = 1- P(not A AND not B) = 1 - P(not A)* P(not B) = 1 - 4/5*3/5 = 13/25
What am I doing wrong? What's the difference with this problem, in which it works:
What is the probability of NOT getting 1 in at least one two independent dice rolls:
Textbook solution: P= P(1) + P(1) + P (1 and 1) = 1/6 + 1/6 - 1/36 = 11/36
Alternative: P= 1 - P(not 1)* P(not 1)= 1- 5/6 * 5/6 = 11/36
03 Jul 2023, 07:59
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