Simultaneous equation with probability. Please help.?

A)
Let P(spin "3") = y

She is 3 times as likely to spin a 1 as to spin a 3.
Then, P(spin "1") = 3y

She is twice as likely to spin an even number as an odd number.
P(spin "2") + P(spin"4") = 2 [P(spin "1") + P(spin "3")]
P(spin "2") + P(spin"4") = 2 [3y + y]
P(spin "2") + P(spin"4") = 8y ...... (1)

The sum of probabilities = 1
P(spin "1") + P(spin"2") + P(spin "3") + P(spin "4") = 1
3y + P(spin"2") + y + P(spin "4") = 1
P(spin "2") + P(spin"4") = 1 - 4y ...... (2)

(1) = (2) :
8y = 1 - 4y
12y = 1
y = 1/12

Hence, P(spin "3") = 1/12


B)
P(spin an odd number)
= P(spin "1") + P(spin "3")
= (1/12) + 3 × (1/12)
= 1/3

P(spin an even number)
= 2 × (1/3)
= 2/3

P(one even number and one odd number)
= P(even number and then odd number) + P(odd number and then even number)
= (2/3) × (1/3) + (1/3) × (2/3)
= 4/9

　
She is twice as likely to spin an even number as an odd number.
P(even) = 2x, P(odd) = x
2x + x = 1
x = 1/3

P(even) = 2/3
P(odd) = 1/3

She is 3 times as likely to spin a 1 as to spin a 3
P(1) = 3y, P(3) = y
3y + y = 1/3
y = 1/12

P(1) = 1/4
P(3) = 1/12
P(2 or 4) = 2/3

A)

What is the probability that she spins a 3?

P(3) = 1/12

B)

What is the probability that, in two spins, she gets one even number and one odd number?

P(even,odd) + P(odd,even)
= 2 (2/3) (1/3)
= 4/9

Well,

let pk = Pr(Anya spins a k)
then we have
p1 = 3 * p3
and
p2 + p4 = 2(p1 + p3) (**)
then
A)
as we also have
p1 + p2 + p3 + p4 = 1
from (**) we obtain
p1 + p3 + 2(p1 + p3) = 1
3(p1 + p3) = 1
p1 + p3 = 1/3
3p3 + p3 = 1/3
4p3 = 1/3
p3 = 1/12
conclusion :
P(Anya spins a 3) = 1/12

B)
P(even) + P(odd) = 1
and
P(even) = 2*P(odd)
therefore
P(odd) = 1/3
and
P(even) = 2/3
and so
P(even and then odd) + P(odd and then even) = 2/9 + 2/9 = 4/9

hope it' ll help !!

There are 4 possible outcomes, each with a different probability, but we know the sum of all four outcomes has to be 1, so let's declare:

Let a = probability of spinning a 1
Let b = probability of spinning a 2
Let c = probability of spinning a 3
Let d = probability of spinning a 4

a + b + c + d = 1

Now that was all before we read anything more than the first sentence. Now we'll read on: 3 times as likely to spin a 1 as a 3, so:

a = 3c

Twice as likely to spin an even number as an odd number, so:

b + d = 2(a + c)

So we have three equations and four unknowns, but I think the b and d will substitute out together. Rearranging the first equation we get:

(b + d) + (a + c) = 1

So we can substitute the expression for (b + d) in terms of a and c to get:

2(a + c) + (a + c) = 1

And simplify:

2a + c + a + c = 1
3a + 2c = 1

Recall that: a = 3c

so now we have:

3(3c) + 2c = 1
9c + 2c = 1
11c = 1
c = 1/11

therefore:

a = 3c
a = 3(1/11)
a = 3/11

So we have the probability of spinning a 1 or a 3 (an odd number) to be the sum of (a + b) , which is 4/11

Therefore, probability of (b + d, even numbers) equals 7/11.

Now that we have that, the probability of getting an even number and an odd number in two spins are the products of the two individual probabilities:

7/11 * 4/11 = 28/121 ≈ 0.231

Recall that there are four outcomes : EE, EO, OE, OO

So we've only computed the EO case. Now we compute the OE case and add it.

The math works out to be the same, so we just double what we computed, so the final probability is:

2 * 28/121 = 56/121 ≈ 0.463

If this was a fair spinner, we'd have calculated this to be 0.5, but since it favors evens to odds, it is skewed a little.