An integer is chosen at random from numbers 1 to 50, what is the probability that the integer chosen is a...
$begingroup$
I considered that multiples of 2 are represented by A, multiples of 3 by B and multiples of 12 by C.
Then I found Union of these 3 sets, which came out to be 29.
I guess I am doing something wrong here only.
Then I found the probability which came out to be 29/50 as total favourable outcomes are 50.
Please someone correct me if I am doing something wrong.
probability
edited Jan 10 at 4:24
Theo Bendit
17.3k12149
asked Jan 10 at 4:17
GarimaGarima
11
$endgroup$
add a comment |
$begingroup$
I considered that multiples of 2 are represented by A, multiples of 3 by B and multiples of 12 by C.
Then I found Union of these 3 sets, which came out to be 29.
I guess I am doing something wrong here only.
Then I found the probability which came out to be 29/50 as total favourable outcomes are 50.
Please someone correct me if I am doing something wrong.
probability
edited Jan 10 at 4:24
Theo Bendit
17.3k12149
asked Jan 10 at 4:17
GarimaGarima
11
$endgroup$
$begingroup$
can you show me how do you obtain $29$?
$endgroup$
– Siong Thye Goh
Jan 10 at 4:26
$begingroup$
Just ignore the multiples of $12$. Every multiple of $12$ is automatically a multiple of $2$ and $3$. Just do an inclusion-exclusion count of the multiples of $2$ and $3$.
$endgroup$
– Theo Bendit
Jan 10 at 4:29
add a comment |
$begingroup$
I considered that multiples of 2 are represented by A, multiples of 3 by B and multiples of 12 by C.
Then I found Union of these 3 sets, which came out to be 29.
I guess I am doing something wrong here only.
Then I found the probability which came out to be 29/50 as total favourable outcomes are 50.
Please someone correct me if I am doing something wrong.
probability
edited Jan 10 at 4:24
Theo Bendit
17.3k12149
asked Jan 10 at 4:17
GarimaGarima
11
$endgroup$
I considered that multiples of 2 are represented by A, multiples of 3 by B and multiples of 12 by C.
Then I found Union of these 3 sets, which came out to be 29.
I guess I am doing something wrong here only.
Then I found the probability which came out to be 29/50 as total favourable outcomes are 50.
Please someone correct me if I am doing something wrong.
probability
probability
edited Jan 10 at 4:24
Theo Bendit
17.3k12149
asked Jan 10 at 4:17
GarimaGarima
11
edited Jan 10 at 4:24
Theo Bendit
17.3k12149
asked Jan 10 at 4:17
GarimaGarima
11
edited Jan 10 at 4:24
Theo Bendit
17.3k12149
edited Jan 10 at 4:24
Theo Bendit
17.3k12149
edited Jan 10 at 4:24
Theo Bendit
17.3k12149
17.3k12149
asked Jan 10 at 4:17
GarimaGarima
11
asked Jan 10 at 4:17
GarimaGarima
11
asked Jan 10 at 4:17
GarimaGarima
11
11
$begingroup$
can you show me how do you obtain $29$?
$endgroup$
– Siong Thye Goh
Jan 10 at 4:26
$begingroup$
Just ignore the multiples of $12$. Every multiple of $12$ is automatically a multiple of $2$ and $3$. Just do an inclusion-exclusion count of the multiples of $2$ and $3$.
$endgroup$
– Theo Bendit
Jan 10 at 4:29
add a comment |
$begingroup$
can you show me how do you obtain $29$?
$endgroup$
– Siong Thye Goh
Jan 10 at 4:26
$begingroup$
Just ignore the multiples of $12$. Every multiple of $12$ is automatically a multiple of $2$ and $3$. Just do an inclusion-exclusion count of the multiples of $2$ and $3$.
$endgroup$
– Theo Bendit
Jan 10 at 4:29
$begingroup$
can you show me how do you obtain $29$?
$endgroup$
– Siong Thye Goh
Jan 10 at 4:26
$begingroup$
can you show me how do you obtain $29$?
$endgroup$
– Siong Thye Goh
Jan 10 at 4:26
$begingroup$
Just ignore the multiples of $12$. Every multiple of $12$ is automatically a multiple of $2$ and $3$. Just do an inclusion-exclusion count of the multiples of $2$ and $3$.
$endgroup$
– Theo Bendit
Jan 10 at 4:29
$begingroup$
Just ignore the multiples of $12$. Every multiple of $12$ is automatically a multiple of $2$ and $3$. Just do an inclusion-exclusion count of the multiples of $2$ and $3$.
$endgroup$
– Theo Bendit
Jan 10 at 4:29
add a comment |
1 Answer
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The set of multiples of 2 and the set of multiples of 3 intersect at the set of multiples of 6 (which contains the set of multiples of 12).
Let $A$ be the set of multiples of 2 in ${1,ldots, 50}$, and $B$ the set of multiples of 3 in ${1,ldots, 50}$. As the set of multiples of 12 in ${1,ldots, 50}$ is in $A cap B$, the set of integers in ${1,2,ldots, 50}$ that is a muliple of 2, 3 or 12 is $A cup B$.
So let us now calculate $|A cup B|$. Note that $|A cup B| = |A| + |B| - |A cap B|$. Note that $A cap B$ the set of multiples of 6 in ${1,ldots, 50}$, while $|A| = 25$ and $|B| = 16$ [make sure you see why], and $|A cap B|$ is 8. So $|A cup B| = 25 + 16 - 8 = 33$.
So the probability that a number drawn from ${1,ldots, 50}$ according to the uniform distribution is a multiple of 2, 3, or 12 is $frac{|A cup B|}{50} = frac{33}{50}$
answered Jan 10 at 4:32
MikeMike
3,492411
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1 Answer
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$begingroup$
The set of multiples of 2 and the set of multiples of 3 intersect at the set of multiples of 6 (which contains the set of multiples of 12).
Let $A$ be the set of multiples of 2 in ${1,ldots, 50}$, and $B$ the set of multiples of 3 in ${1,ldots, 50}$. As the set of multiples of 12 in ${1,ldots, 50}$ is in $A cap B$, the set of integers in ${1,2,ldots, 50}$ that is a muliple of 2, 3 or 12 is $A cup B$.
So let us now calculate $|A cup B|$. Note that $|A cup B| = |A| + |B| - |A cap B|$. Note that $A cap B$ the set of multiples of 6 in ${1,ldots, 50}$, while $|A| = 25$ and $|B| = 16$ [make sure you see why], and $|A cap B|$ is 8. So $|A cup B| = 25 + 16 - 8 = 33$.
So the probability that a number drawn from ${1,ldots, 50}$ according to the uniform distribution is a multiple of 2, 3, or 12 is $frac{|A cup B|}{50} = frac{33}{50}$
answered Jan 10 at 4:32
MikeMike
3,492411
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add a comment |
$begingroup$
The set of multiples of 2 and the set of multiples of 3 intersect at the set of multiples of 6 (which contains the set of multiples of 12).
Let $A$ be the set of multiples of 2 in ${1,ldots, 50}$, and $B$ the set of multiples of 3 in ${1,ldots, 50}$. As the set of multiples of 12 in ${1,ldots, 50}$ is in $A cap B$, the set of integers in ${1,2,ldots, 50}$ that is a muliple of 2, 3 or 12 is $A cup B$.
So let us now calculate $|A cup B|$. Note that $|A cup B| = |A| + |B| - |A cap B|$. Note that $A cap B$ the set of multiples of 6 in ${1,ldots, 50}$, while $|A| = 25$ and $|B| = 16$ [make sure you see why], and $|A cap B|$ is 8. So $|A cup B| = 25 + 16 - 8 = 33$.
So the probability that a number drawn from ${1,ldots, 50}$ according to the uniform distribution is a multiple of 2, 3, or 12 is $frac{|A cup B|}{50} = frac{33}{50}$
answered Jan 10 at 4:32
MikeMike
3,492411
$endgroup$
add a comment |
$begingroup$
The set of multiples of 2 and the set of multiples of 3 intersect at the set of multiples of 6 (which contains the set of multiples of 12).
Let $A$ be the set of multiples of 2 in ${1,ldots, 50}$, and $B$ the set of multiples of 3 in ${1,ldots, 50}$. As the set of multiples of 12 in ${1,ldots, 50}$ is in $A cap B$, the set of integers in ${1,2,ldots, 50}$ that is a muliple of 2, 3 or 12 is $A cup B$.
So let us now calculate $|A cup B|$. Note that $|A cup B| = |A| + |B| - |A cap B|$. Note that $A cap B$ the set of multiples of 6 in ${1,ldots, 50}$, while $|A| = 25$ and $|B| = 16$ [make sure you see why], and $|A cap B|$ is 8. So $|A cup B| = 25 + 16 - 8 = 33$.
So the probability that a number drawn from ${1,ldots, 50}$ according to the uniform distribution is a multiple of 2, 3, or 12 is $frac{|A cup B|}{50} = frac{33}{50}$
answered Jan 10 at 4:32
MikeMike
3,492411
$endgroup$
The set of multiples of 2 and the set of multiples of 3 intersect at the set of multiples of 6 (which contains the set of multiples of 12).
Let $A$ be the set of multiples of 2 in ${1,ldots, 50}$, and $B$ the set of multiples of 3 in ${1,ldots, 50}$. As the set of multiples of 12 in ${1,ldots, 50}$ is in $A cap B$, the set of integers in ${1,2,ldots, 50}$ that is a muliple of 2, 3 or 12 is $A cup B$.
So let us now calculate $|A cup B|$. Note that $|A cup B| = |A| + |B| - |A cap B|$. Note that $A cap B$ the set of multiples of 6 in ${1,ldots, 50}$, while $|A| = 25$ and $|B| = 16$ [make sure you see why], and $|A cap B|$ is 8. So $|A cup B| = 25 + 16 - 8 = 33$.
So the probability that a number drawn from ${1,ldots, 50}$ according to the uniform distribution is a multiple of 2, 3, or 12 is $frac{|A cup B|}{50} = frac{33}{50}$
answered Jan 10 at 4:32
MikeMike
3,492411
answered Jan 10 at 4:32
MikeMike
3,492411
answered Jan 10 at 4:32
MikeMike
3,492411
answered Jan 10 at 4:32
MikeMike
3,492411
3,492411
add a comment |
add a comment |
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$begingroup$
can you show me how do you obtain $29$?
$endgroup$
– Siong Thye Goh
Jan 10 at 4:26
$begingroup$
Just ignore the multiples of $12$. Every multiple of $12$ is automatically a multiple of $2$ and $3$. Just do an inclusion-exclusion count of the multiples of $2$ and $3$.
$endgroup$
– Theo Bendit
Jan 10 at 4:29