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Probability of one event or another with 2 spinners

2

$begingroup$

Problem: First spinner has 4, 2, 3 with 4 as 1/2 the spinner, and 2 and 3 each 1/4 of spinner. Second spinner has 5 and 6 each as 1/2 the spinner. Find the P(first spin is an even number OR the sums of the two spins is 9).

I made a tree diagram with sample space and probablities. I know probablities for these: (4,5) and (4,6) is 1/4 each; and (2,5)(2,6)(3,5)and (3,6) each have probability of 1/8. I believe P(even on first spin) = 1/2 + 1/4 = 3/4. And P(the sum of two spins is 9) = 1/4 + 1/8 = 3/8.

To find P(even on first spin OR the sums of the two spins is 9) do you take P(even on first) + P(the sum of two spins is 9) - P(even on first AND sum of two spins is 9)? So 3/4 + 3/8 - (1/4) = 7/8? I feel like I am doing wrong and hope someone can explain. Thanks!

probability

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asked Apr 1 at 18:00

Julie JJulie J

154

$endgroup$

add a comment | 

2

$begingroup$

Problem: First spinner has 4, 2, 3 with 4 as 1/2 the spinner, and 2 and 3 each 1/4 of spinner. Second spinner has 5 and 6 each as 1/2 the spinner. Find the P(first spin is an even number OR the sums of the two spins is 9).

I made a tree diagram with sample space and probablities. I know probablities for these: (4,5) and (4,6) is 1/4 each; and (2,5)(2,6)(3,5)and (3,6) each have probability of 1/8. I believe P(even on first spin) = 1/2 + 1/4 = 3/4. And P(the sum of two spins is 9) = 1/4 + 1/8 = 3/8.

To find P(even on first spin OR the sums of the two spins is 9) do you take P(even on first) + P(the sum of two spins is 9) - P(even on first AND sum of two spins is 9)? So 3/4 + 3/8 - (1/4) = 7/8? I feel like I am doing wrong and hope someone can explain. Thanks!

probability

share|cite|improve this question

asked Apr 1 at 18:00

Julie JJulie J

154

$endgroup$

add a comment | 

$begingroup$

Problem: First spinner has 4, 2, 3 with 4 as 1/2 the spinner, and 2 and 3 each 1/4 of spinner. Second spinner has 5 and 6 each as 1/2 the spinner. Find the P(first spin is an even number OR the sums of the two spins is 9).

I made a tree diagram with sample space and probablities. I know probablities for these: (4,5) and (4,6) is 1/4 each; and (2,5)(2,6)(3,5)and (3,6) each have probability of 1/8. I believe P(even on first spin) = 1/2 + 1/4 = 3/4. And P(the sum of two spins is 9) = 1/4 + 1/8 = 3/8.

To find P(even on first spin OR the sums of the two spins is 9) do you take P(even on first) + P(the sum of two spins is 9) - P(even on first AND sum of two spins is 9)? So 3/4 + 3/8 - (1/4) = 7/8? I feel like I am doing wrong and hope someone can explain. Thanks!

probability

share|cite|improve this question

asked Apr 1 at 18:00

Julie JJulie J

154

$endgroup$

share|cite|improve this question

asked Apr 1 at 18:00

Julie JJulie J

154

share|cite|improve this question

asked Apr 1 at 18:00

Julie JJulie J

154

asked Apr 1 at 18:00

Julie JJulie J

154

asked Apr 1 at 18:00

Julie JJulie J

154

1 Answer
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1

$begingroup$

I see nothing wrong with your argument.

But since you have all the outcomes and their respective probabilities written out already, you can also count directly. The outcomes which satisfy "first spinner is even or sum is 9" are (4,5), (4,6), (2,5), (2,6) and (3,6) for a total probability of
$$
frac14+frac14+frac18+frac18+frac18=frac78
$$
Alternatively, the single outcome which doesn't satisfy satisfy this is (3,5) with probability $frac18$, which also gives the same answer.

share|cite|improve this answer

edited Apr 1 at 18:17

answered Apr 1 at 18:06

ArthurArthur

123k7122211

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1

$begingroup$

I see nothing wrong with your argument.

But since you have all the outcomes and their respective probabilities written out already, you can also count directly. The outcomes which satisfy "first spinner is even or sum is 9" are (4,5), (4,6), (2,5), (2,6) and (3,6) for a total probability of
$$
frac14+frac14+frac18+frac18+frac18=frac78
$$
Alternatively, the single outcome which doesn't satisfy satisfy this is (3,5) with probability $frac18$, which also gives the same answer.

share|cite|improve this answer

edited Apr 1 at 18:17

answered Apr 1 at 18:06

ArthurArthur

123k7122211

$endgroup$

add a comment | 

1

$begingroup$

I see nothing wrong with your argument.

But since you have all the outcomes and their respective probabilities written out already, you can also count directly. The outcomes which satisfy "first spinner is even or sum is 9" are (4,5), (4,6), (2,5), (2,6) and (3,6) for a total probability of
$$
frac14+frac14+frac18+frac18+frac18=frac78
$$
Alternatively, the single outcome which doesn't satisfy satisfy this is (3,5) with probability $frac18$, which also gives the same answer.

share|cite|improve this answer

edited Apr 1 at 18:17

answered Apr 1 at 18:06

ArthurArthur

123k7122211

$endgroup$

add a comment | 

$begingroup$

I see nothing wrong with your argument.

But since you have all the outcomes and their respective probabilities written out already, you can also count directly. The outcomes which satisfy "first spinner is even or sum is 9" are (4,5), (4,6), (2,5), (2,6) and (3,6) for a total probability of
$$
frac14+frac14+frac18+frac18+frac18=frac78
$$
Alternatively, the single outcome which doesn't satisfy satisfy this is (3,5) with probability $frac18$, which also gives the same answer.

share|cite|improve this answer

edited Apr 1 at 18:17

answered Apr 1 at 18:06

ArthurArthur

123k7122211

$endgroup$

share|cite|improve this answer

edited Apr 1 at 18:17

answered Apr 1 at 18:06

ArthurArthur

123k7122211

answered Apr 1 at 18:06

ArthurArthur

123k7122211

answered Apr 1 at 18:06

ArthurArthur

123k7122211