1. ## Probability Spinner Question?

I am confused on how to answer this question, I know its simple as I've done it before but I'm having a mental block.

A five-sided spinner's sides are labelled 1,2,3,4,5. The spinner is biased. The probability that the spinner will land on each number is given:

Probability: (1) - 0.36 (2) - 0.1 (3) - 0.25 (4) - 0.15 (5) - 0.14

a) Chris spins the spinner 50 times, work out an estimate for the number of times the spinner lands on 1.

b) Dylan spins the spinner twice. Work out the probability that:

i) it lands on 2 both times.
ii) the spinner will land on an even number exactly once.

Help would be appreciated

2. Originally Posted by abbie1212
I am confused on how to answer this question, I know its simple as I've done it before but I'm having a mental block.

A five-sided spinner's sides are labelled 1,2,3,4,5. The spinner is biased. The probability that the spinner will land on each number is given:

Probability: (1) - 0.36 (2) - 0.1 (3) - 0.25 (4) - 0.15 (5) - 0.14

a) Chris spins the spinner 50 times, work out an estimate for the number of times the spinner lands on 1.

b) Dylan spins the spinner twice. Work out the probability that:

i) it lands on 2 both times.
ii) the spinner will land on an even number exactly once.

Help would be appreciated
it will be helpful if you labelled each case from 1-5:
let P(x) denote the probability that a spin lands on x, (x:1-5)

a) 50 times, each time it lands on 1 with P(1)=0.36, hence 50*P(1)=18
b)
i) simply P(2)*P(2)=0.01
ii) this is a bit tricky, but with a bit of logical thinking, you can figure out that: p=P(even)*P(odd)+P(odd)*P(even)=2*P(even)*P(odd)=2 *(P(2)+P(4))*(P(1)+P(3)+P(5))

I'll leave the last simple calculation for you to finish

3. I don't really understand the logic behind b ii

I would just calculate the odds that you roll odd (which is 3/4)
and then calculate the odds that you roll even (which is 1/4)

So, the odds of rolling 2 odds in a row are 3/8 (3/4*3/4) and the odds of rolling 2 evens in a row are 1/8 (1/4*1/4)

then you add the odds of rolling 2 evens plus the odds of rolling 2 odds together and you get 1/2 (1/8+3/8=4/8) so, you have a 50/50 shot of rolling one even.

Another, much simpler way to look at it is this:
0.1+0.15=0.25 so, you have a 1/4 shot to roll a an even. if you get 2 rolls to try to roll an even once, you just double your odds and get .5 (.25+.25)

Another way to look at it would be this: first calculate all the discrete combinations

1,1
1,2
1,3
1,4
1,5
2,1
2,3
2,4
2,5
3,1
3,2
3,3
3,4
3,5
4,1
4,2
4,3
4,4
4,5
5,1
5,2
5,3
5,4
5,5
Then you multiply each of the chances so 1,1 is .36*.36 and 1,3 is .36*.25 and you go on and do that for all 25 possible combinations, then you take all the answers that have one odd and one even and add them together, and you take all the ones that are two odds or two evens and you add them together too. I haven't bothered to do it, but unless I am wrong in the first two examples, they would both add up to .5 as well

4. Doing far too much work for part ii:

Each of the five spinners are mutually exclusive events (no two can occur at the same time). Therefore the probability of an even on one spin is simply the sum of the probabilities of 2 and 4. One can get an even exactly once two ways - even, odd and odd,even. Thus:

P(Even exactly once)=(E,O)+(O,E).

5. Originally Posted by abbie1212
A five-sided spinner's sides are labelled 1,2,3,4,5. The spinner is biased. The probability that the spinner will land on each number is given:
Probability: (1) - 0.36 (2) - 0.1 (3) - 0.25 (4) - 0.15 (5) - 0.14
b) Dylan spins the spinner twice. Work out the probability that:
ii) the spinner will land on an even number exactly once.
Originally Posted by papakapp
I don't really understand the logic behind b ii
Don’t make it more than it is.
On any spin the probability of an even outcome is $\displaystyle 0.25$ while the probability of an odd outcome is $\displaystyle 0.75$.
So on two spins the probability of exactly one even outcome is $\displaystyle 2(0.25)(0.75)=0.375$.

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# number of times biased spinner lands on 1

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