## mathmath333 one year ago probability question

1. mathmath333

\large \color{black}{\begin{align} & \normalsize \text{Probability of solving specific problem independently by A and B are}\hspace{.33em}\\~\\ & \dfrac12\ \ \normalsize \text{and}\ \ \dfrac13\ \normalsize \text{respectively.}\hspace{.33em}\\~\\ & \normalsize \text{If both try to solve the problem independently, find the probability}\hspace{.33em}\\~\\ & \normalsize \text{ that exactly one of them solves the problem.}\hspace{.33em}\\~\\ \end{align}}

2. ganeshie8

you want to find below two probabilities and then add them : 1) A solves the problem but B doesn't 2) A doesn't solve the problem but B solves it

3. ganeshie8

P("A solves" AND "B doesnt solve") = ?

4. mathmath333

how did u know i need to find this 1) A solves the problem but B doesn't 2) A doesn't solve the problem but B solves it

5. ganeshie8

good question, thats because of the phrase "exactly one of them"

6. ganeshie8

"exactly one of them" means only one of them solves the problem

7. mathmath333

ok

8. mathmath333

P("A solves" AND "B doesnt solve") = P(A$$\cap$$B')

9. mathmath333

is this correct P("A solves" AND "B doesnt solve") = P(A∩B')

10. phi

yes

11. ganeshie8

Correct, it is easy, go ahead and find it...

12. mathmath333

P(A∩B')=P(A)-P(A∩B)

13. ganeshie8

i suggest you not use the formulas to solve these problems, use the problem to make sense of formulas instead

14. thomas5267

There are only four situations. 1. A solves it and B couldn't. 2. A couldn't solve it and B solves it. 3. A and B both solve it. 4. A and B both couldn't solve it (idiots!) The question is asking for the probability of exactly one of them solves it (or one of them is an idiot!), so it is asking for the probability of 1 or 2 happening.

15. mathmath333

how can i solve it without formula "P(A∩B')"

16. ganeshie8

you're given, probability that B solves the problem = 1/3 so can you guess the probability that B couldn't solve the problem ?

17. mathmath333

the probability that B couldn't solve the problem=2/3

18. thomas5267

I think it is true for all events X, P(X')=1-P(X).

19. ganeshie8

Yes, since the events are independent, simply multiply the probabilities : P("A solves" AND "B doesnt solve") = P("A solves")*P("B doesnt solve") = ?

20. mathmath333

the probability that B couldn't solve the problem =1/2*2/3=1/3

21. thomas5267

Put it in another way, an event can either happen or not happen. No event can not happen and not not happen at the same time.

22. ganeshie8

Correct. try finding the probability for other case too : 2) A doesn't solve the problem but B solves it

23. mathmath333

P("A doesnt solves" AND "B does solve") =1/6

24. ganeshie8

Add them up and you're done!

25. mathmath333

=1/2

26. ganeshie8

1/3 + 1/6 = 3/6 yeah 1/2 looks good to me

27. mathmath333

thnx