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artcher [175]
2 years ago
9

A number is chosen at random from the positive, even integers from 2 to 50. Find the probability that the number chosen is divis

ible by 5 or 8.
1/25

2/5

11/25
Mathematics
1 answer:
VashaNatasha [74]2 years ago
3 0
There are 25 even integers between 2 and 50 (2,4,6,...,50).
Among them, there are 5 numbers divisible by 5 (10,20,30,40,50), 6 divisible by 8 (8,16,24,32,40,48), and 1 divisible by 5 and 8 simultaneously (40).

The last means these two events are not mutually exclusive. So, if

A - the event of choosing a number divisible by 5,
B - the event of choosing a number divisible by 8,

the probability of choosing a number divisible by 5 or 8 is equal to
P(A\cup B)=P(A)+P(B)-P(A\cap B)

So,
P(A\cup B)=\dfrac{5}{25}+\dfrac{6}{25}-\dfrac{1}{25}\\
P(A\cup B)=\dfrac{10}{25}\\
\boxed{P(A\cup B)=\dfrac{2}{5}}


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<u></u>

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Explanation:

The figure attached shows the <em>Venn diagram </em>for the given sets.

<em />

<em><u>a) What is the probability that the number chosen is a multiple of 3 given that it is a factor of 24?</u></em>

<em />

From the whole numbers 1 to 15, the multiples of 3 that are factors of 24 are in the intersection of the two sets: 3, 6, and 12.

There are a total of 7 multiples of 24, from 1 to 15.

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<em><u /></em>

<em><u>b) What is the probability that the number chosen is a factor of 24 given that it is a multiple of 3?</u></em>

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