Probability question help!

- Assume 60% of population have DVD and 35% have iPod. If someone is selected at random what is probability that they either have a DVD or iPod?
- In a factory, machines of type A produce 30% of the output, machines of type B produce 25%, and machines of type C produce 45% of the output. On average, 2% of the produce from the machines of type A is defective, with 3% defective from type B and 4% from type C. Calculate the probability that an item of output from the factory chose at random is defective. If a defective item is chosen calculate the probability that it was made by a machine of type C?
- The police discover a murder victim at 6:00am. The body temp of victim is measured then found to 28 deg C. A doctor arrives on the scene of the crime 60 minutes later and measures the body temp again. It is found to be 24 deg C. the temp of the room remains constant at 12 deg C. the doctor knowing that normal body temp is 37 deg C, is able to estimate the time of death of the victim. If the cooling process is modelled by the equation θ = θ
_{0 }е^{-kt}, where θ is the excess temp over room temperature, use this model to estimate the time of death.

Re: Probability question help!

Re: Probability question help!

explain how you got to this answer

Re: Probability question help!

Quote:

Originally Posted by

**Rizwan123** - Assume 60% of population have DVD and 35% have iPod. If someone is selected at random what is probability that they either have a DVD or iPod?

That question is so poorly worded that I doubt that it can be answered. We do not know if the two groups are disjoint. If not, we don’t know how they intersect. Therefore,

Quote:

Originally Posted by

**wnvl** 1) 0.6*0.65+0.35*0.4

has no meaning whatsoever.

Re: Probability question help!

they are disjoint,

could you please explain how to do it. what about 2 and 3

Re: Probability question help!

Quote:

Originally Posted by

**Rizwan123** they are disjoint,

**Why was that not stated in the OP?**

So $\displaystyle 0.05$ have neither.

Thus $\displaystyle 1-0.5$ have at least one.

Re: Probability question help!

that answer has no meaning, explain how you arised to it.

is there a way of wolfram alpha calculating it?

Re: Probability question help!

Quote:

Originally Posted by

**wnvl** 1) 0.6*0.65+0.35*0.4

=P(person has no IPOD)*P(person has DVD)+P(person has IPOD)*P(peson has no DVD)

clear now?

I assume you mean people that have a DVD or an IPOD and not both.