put on 5 numbers onto the spinner so that:

the probability of spinning a three is 1/2 and
the probability of spinning an evn number is 1/3

(there are 6 faces on the spinner and there is already a number 1 , witch makes out that only 5 faces are blank and need filling in)

what are they?

thanks

Quote:

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Originally Posted by awilliams666

put on 5 numbers onto the spinner so that:

the probability of spinning a three is 1/2 and
the probability of spinning an evn number is 1/3

(there are 6 faces on the spinner and there is allready a number 1 , witch makes out that only 5 faces are blank and need filling in)

what are they?

thanks

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So if there are six spaces and the probability of spinning a 3 is 1/2 then half the spaces need to be a 3.

This will leave only 2 spaces left that don't have a number since the question states that one out of the six spaces already has a number 1.

Since there is a 1/3 probability of spinning an even number both of the remaining spaces need to be any even number.

Iamtheman,

We don't do people's homework in this website. If the OP posted a partially completed problem or was getting stuck in one specific spot that would be different, but then again we only help guide them in the right direction.

Lowtax4eva,

You may think that I'm giving the answer away but keep in mind the question that is being asked. If awilliam is asking this question then the answer I gave certainly does not give the answer away.

Please refrain from replying on a subject which you clearly don't know a whole lot about.

Thanks!