1. ## binomial probability

There are 5 trials each with success probability of 0.02. The probab of winning only at the 3rd AND 5th trial is 0.02^2 * 0.98^3.

What if I am asked the prob of winning at the 3rd OR at the 5th trial?

How do I do it? And why?

2. Originally Posted by 0123
There are 5 trials each with success probability of 0.02. The probab of winning only at the 3rd AND 5th trial is 0.02^2 * 0.98^3.

What if I am asked the prob of winning at the 3rd OR at the 5th trial?

How do I do it? And why?
One way of doing this is to observe that:

P(3 or 5)= P(3 but not 5) + P(3 and 5) + P(5 but not 3)=0.02*0.98 + 0.02*0.02 + 0.98*0.02

this is assuming you are indifferent to the results of the other trials.

RonL

3. Originally Posted by CaptainBlack
One way of doing this is to observe that:

P(3 or 5)= P(3 but not 5) + P(3 and 5) + P(5 but not 3)=0.02*0.98 + 0.02*0.02 + 0.98*0.02

this is assuming you are indifferent to the results of the other trials.

RonL
What does it mean that I am indifferent to the other trials? And why this time we do not do like before I mean multiplying for the times we get it wrong like 0.02^2* 0.98^3 ? Thank you so so much Captain Black

4. Originally Posted by 0123
There are 5 trials each with success probability of 0.02. The probability of winning only at the 3rd AND 5th trial is 0.02^2 * 0.98^3.
What if I am asked the prob of winning at the 3rd OR at the 5th trial?
How do I do it? And why?
Do you mean “winning at exactly either the third or at the fifth and not both”?
Do you mean “winning at least the third or at the fifth”?
Do you mean “winning at only the third or at the fifth”?
Or what do you mean?

5. Originally Posted by Plato
Do you mean “winning at exactly either the third or at the fifth and not both”?
Do you mean “winning at least the third or at the fifth”?
Do you mean “winning at only the third or at the fifth”?
Or what do you mean?
I am doing this stuff for my statistics course and there is this exercice:

The probability of winning when playing one game on the slot machine is 2%. Mr Brown tries 5 times(assume independently). What is the probability of winning only at the third and the fifth trials?

That is the exercice and in class we solved it like this: 0.02^2* 0.98^3.
In class we never seen the "OR" and I was wondering what I am supposed to do if at the exam they put me an exercice with "Or" istead of "and". Can you help me? Can you explaining me what all your questions mean? I mean can you let me see how they are different? Thank you so so much.

6. Originally Posted by 0123
I am doing this stuff for my statistics course and there is this exercice:
The probability of winning when playing one game on the slot machine is 2%. Mr Brown tries 5 times(assume independently). What is the probability of winning only at the third and the fifth trials?.
If you change “What is the probability of winning only at the third and the fifth trials” to “What is the probability of winning only at the third OR the fifth trials, I would take that to mean “winning at the third or at the fifth but not both”. That would be $\left( {0.02} \right)\left( {0.98} \right)^4 + \left( {0.02} \right)\left( {0.98} \right)^4 = 2\left( {0.02} \right)\left( {0.98} \right)^4$.

But you really need to discuss this with your instructor. Find out her/his take on the language of this question.

7. well I was just trying to figure out what to do in case I find an "OR" somewhere. We are not supposed to hae the same exercices we did in class. They are usually things we have never seen before, that is why I was wondering about the or.

So to wrap up: if I find something that appears to mean "3rd(let's say) or 5th but not both" I compute the prob of getting ONLY the 3rd as success just thinking the others are all failures. do the same with the 5th and sum them.

If I find "3rd or 5th or both" the same as before plus the probab of having exacltly those two successes.

Is this right? Are there any other cases I might find? Might you please let me see how to answer the questions you posed, I have no idea where to start?
Thanks a lot

8. Originally Posted by 0123
If I find "3rd or 5th or both" the same as before plus the probab of having exacltly those two successes.
I would expect my students to answer: $\left( {0.02} \right)\left( {0.98} \right)^4 + \left( {0.02} \right)\left( {0.98} \right)^4 + \left( {0.02} \right)^2 \left( {0.98} \right)^3$.
That is only at 3 plus only at 5 plus only at 3&5.

The word “only” is the trouble here. And when used with ‘or’ double-trouble.