problem with probability

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pipo

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Apr 9, 2020
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Hello everyone
I am new to the statistics. I am not very good, unfortunatelly. I have to solve some task like a homework. The problem is we didn't have this topic in class.
Can I have some of your help, please ? Here is the tasks :
In a contest, Chris, Albert, and John, are asked to solve a problem, the respective probabilities that they solve the problem are 1/6, 1/8, 1/3.
Calculate the probability that:
1. None of them solves the problem.
2. At least one of them solves the problem.
3. Only one of them solves the problem.
I tried myself for 1. and get 99.34%. Is that right ?
But I have no clue about 2. and 3.
Any help ?
Pleeeeese :)
 
1. Your answer is incorrect. Use two facts:

_ i) the problem solving performance is independent over the players,
_ ii) Prob(not A) = 1 - Prob(A).

2. Use ii).

3. Write down all possible scenarios and sum up their probabilities.
 
Here's how I would do that:
Imagine this happening 1440 times.

Charlie will solve the problem 1440/6= 240 times. He will NOT solve it 1440- 240= 1200 times. Of those 1200 times, Albert will solve it 1200/8= 150 times and NOT solve it 1200- 150= 1050 times. Of those 1050 times John will solve it 1050/3= 350 times and not solve it 1050- 350= 700 times. All three will not solve it 700/1440= 0.486 of the time. The probability all three will NOT solve the problem is 0.486 or 48.6%.

We can do the same thing to calculate the probability that exactly one solves the problem. As above Charlie and Albert will NOT solve the problem but John WILL 350 times out of the 1440.
Charlie will not solve the problem but Albert will 150 times and of those 150 times John will solve it 150/3= 50 times so will not solve it 100 times. That is, Charlie and John will not solve the problem and Albert will 100 times out of 1440.
Charlie will solve the problem 240 times. Of those 240 times Albert will solve it 240/8= 30 times and not solve it 240- 30= 210 times. Of those 210 times John will solve it 210/3= 70 and not solve it 210- 70= 140 times. Charlie will solve the problem but Albert and John not solve it 140 times out of 1440 times.
That is, exactly one of the three will solve the problem a total of 350+ 100+ 140= 590 times out of 1440. That is a probability of 590/1440= 0.410 or 41%.

Do the last one the same way.
 
pipo said:
Here is the tasks : In a contest, Chris, Albert, and John, are asked to solve a problem, the respective probabilities that they solve the problem are 1/6, 1/8, 1/3. Calculate the probability that:
1. None of them solves the problem.
2. At least one of them solves the problem.
3. Only one of them solves the problem.
Click to expand...
This is a poorly put question. Therefore let's assume the probabilities are independent.
\(\mathcal{P}{(C)}=\frac{1}{6},~\mathcal{P}{(A)}=\frac{1}{8},~\&~\mathcal{P}{(J)}=\frac{1}{3}\)
If the events are independent so are their complements.
1) Thus none solve is: \(\mathcal{P}(C^c)\cdot\mathcal{P}(A^c)\cdot\mathcal{P}(J^c)\)
2) Is the complement of 1).
3) Exactly one is \(\mathcal{P}(C)\cdot\mathcal{P}(A^c)\cdot\mathcal{P}(J^c)+\mathcal{P}(A)\cdot\mathcal{P}(C^c)\cdot\mathcal{P}(J^c)+\mathcal{P}(J)\cdot\mathcal{P}(A^c)\cdot\mathcal{P}(C^c)\)
 
Andreea Alexandra said:
i have the same problem to solve and i didnt quit understand .can you help me please?
thank you
Click to expand...
Can you explain what part of the given solution do you not understand? Can you share your attempt with us so we know where you are making any mistakes and we can then help you by given you hints?
 
hi.thaknks for ur reply.
so for none of them i got the formula that pka wrote above and the final result is 0.006.im not sure if thats right.
for the second with at least one of them im not sure what the answer is.is it 1?and if yes there is a formula or i just write 1?
and for the 3 do i use the formula above that pka wrote...thats where im not sure either
thank you
 
and i have another if you can help me please.
A box A contains 3 white and 2 blue balls, and another box B contains 4 green and 5 blue balls. If a ball is picked random from each box,find the probability that:
1. one is green and the other is white.
2. they are of the same colour.
for 1 i put that P of green from Box B is 4/9 which is 0.44
P of white from box A is 3/5 which is 0.6
i did 4/9X3/5=0.266 is that correct?
but for the question number 2 i dont know how to do it.
Can you help me please?
Thank you
 
Do you understand what "\(\displaystyle A^c\)", "\(\displaystyle J^c\)", and "\(\displaystyle C^c\)"mean? You are told that \(\displaystyle P(C)= \frac{1}{6}\) so what is \(\displaystyle P(C^c)\)?
 
I think i got it right this time.
none of them is 1-1/6=0.84
1-1/8=0,875
1-1/3=0.67
0.84×0.875×0.67=0.49
For at least one is the complement meaning 1-0.49=0.51
Am i right?
 
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