mathmath333
  • mathmath333
probability question
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{Probability of solving specific problem independently by A and B are}\hspace{.33em}\\~\\ & \dfrac12\ \ \normalsize \text{and}\ \ \dfrac13\ \normalsize \text{respectively.}\hspace{.33em}\\~\\ & \normalsize \text{If both try to solve the problem independently, find the probability}\hspace{.33em}\\~\\ & \normalsize \text{ that exactly one of them solves the problem.}\hspace{.33em}\\~\\ \end{align}}\)
ganeshie8
  • 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
ganeshie8
  • ganeshie8
P("A solves" AND "B doesnt solve") = ?

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mathmath333
  • 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
ganeshie8
  • ganeshie8
good question, thats because of the phrase "exactly one of them"
ganeshie8
  • ganeshie8
"exactly one of them" means only one of them solves the problem
mathmath333
  • mathmath333
ok
mathmath333
  • mathmath333
P("A solves" AND "B doesnt solve") = P(A\(\cap\)B')
mathmath333
  • mathmath333
is this correct P("A solves" AND "B doesnt solve") = P(A∩B')
phi
  • phi
yes
ganeshie8
  • ganeshie8
Correct, it is easy, go ahead and find it...
mathmath333
  • mathmath333
P(A∩B')=P(A)-P(A∩B)
ganeshie8
  • ganeshie8
i suggest you not use the formulas to solve these problems, use the problem to make sense of formulas instead
thomas5267
  • 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.
mathmath333
  • mathmath333
how can i solve it without formula "P(A∩B')"
ganeshie8
  • 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 ?
mathmath333
  • mathmath333
the probability that B couldn't solve the problem=2/3
thomas5267
  • thomas5267
I think it is true for all events X, P(X')=1-P(X).
ganeshie8
  • 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") = ?
mathmath333
  • mathmath333
the probability that B couldn't solve the problem =1/2*2/3=1/3
thomas5267
  • 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.
ganeshie8
  • ganeshie8
Correct. try finding the probability for other case too : 2) A doesn't solve the problem but B solves it
mathmath333
  • mathmath333
P("A doesnt solves" AND "B does solve") =1/6
ganeshie8
  • ganeshie8
Add them up and you're done!
mathmath333
  • mathmath333
=1/2
ganeshie8
  • ganeshie8
1/3 + 1/6 = 3/6 yeah 1/2 looks good to me
mathmath333
  • mathmath333
thnx

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