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tankabanditka [31]
3 years ago
8

Lenny rolls two number cubes. What is the probability that the sum of the numbers Lenny rolls is either 4 or 7?

Mathematics
2 answers:
Mashutka [201]3 years ago
7 0

Probability of an event =


(number of ways the event can happen) / (total number of possible outcomes) .


Well, the total number of possible outcomes is easy. It's


(6 possibilities for the first cube) x (6 possibilities for the 2nd cube)


= 36 total possible ways for two cubes to roll.


Now, how many of those possibilities show either 4 or 7 ?


Well, here's how a 4 can come up:


1 . . . 3

2 . . . 2

3 . . . 1


and here's how a 7 can come up:


1 . . . 6

2 . . . 5

3 . . . 4

4 . . . 3

5 . . . 2

6 . . . 1 .


So there are 9 ways all together for Lenny to be successful.


The probability is (9/36) .


That's 1/4 , or 25% .

WITCHER [35]3 years ago
6 0

|\Omega|=6^2=36\\ |A|=\underbrace{3}_{\text{the sum is 4}}+\underbrace{6}_{\text{the sum is 7}}=9\\\\ P(A)=\dfrac{9}{36}=\dfrac{1}{4}=25\%

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Substitution method:

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2 years ago
If A and B are independent events with P(A) = .5 and P(B) = .2, find the following:a) P(A U B)b) P(A^c ? B^c)c) P(A^c U B^c)**No
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a.

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P(A\cup B)=0.5+0.2-0.5\cdot0.2

\boxed{P(A\cup B)=0.6}

b. I'm guessing the ? is supposed to stand for intersection. We can use DeMorgan's law for complements here:

P(A^c\cap B^c)=P(A\cup B)^c=1-P(A\cup B)

P(A^c\cap B^c)=1-0.6

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P(A^c\cup B^c)=P(A\cap B)^c=1-P(A\cap B)=1-P(A)P(B)

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Answer:

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