box item problem

Printable View

May 16th 2010, 10:37 PM
kingman

box item problem

Dear Sir,
I need help some help with the below problem and would appreciate very much if can offer some help.
thanks
Kingman

John has 5 items, one each to be named P,Q, R,S and T. Each item is placed in a separate
box and sealed. He then addresses the boxes, at random, one each to P,Q,R,S and T.
1) Find the probability that the item P is in the correct box and the letter Q is in the
wrong box. (3/20)
2) Find the probability that the item P is in the correct box, given that the item Q is in the
wrong box.(3/16)
3) Find the probability that both of the items to P and Q are in the wrong boxes.(13/20)

Hint: It would be helpful to draw a Venn diagram showing the various given probabilities.
A suitable tree diagram is also helpful.
Lastly let X denotes the event that X’s item is in the correct box.
May 17th 2010, 07:26 AM
Soroban

Hello, kingman!

I can't imagine how to draw a Venn diagram for this problem.

Quote:

John has 5 items named $P,Q, R,S, T$ .
Each item is placed in a separate box and sealed.
He then addresses the boxes at random, one each to $P, Q, R, S, T.$

1) Find the probability that the item $P$ is in the correct box
and item $Q$ is in the wrong box. (3/20)

First, probability that P is correct: . $\frac{1}{5}$
Then probability that Q is incorrect: . $\frac{3}{4}$

Therefore: . $p\left(P \wedge \overline{Q}\right) \;=\;\frac{1}{5}\cdot\frac{3}{4} \;=\;{\color{blue}\frac{3}{20}}$

Quote:

2) Find the probability that the item $P$ is in the correct box,
given that the item $Q$ is in the wrong box. (3/16)

Bayes' Theorem: . $p\left(P\,|\,\overline{Q}\right) \;=\;\frac{p\left(P \wedge \overline{Q}\right)}{p\left(\overline{Q}\right)}$ .[1]

We found the numerator in part (1): . $p(P\,\wedge\,\overline{Q}) \;=\;\frac{3}{20}$

The denominator is: . $p(\overline{Q}) \:=\:\frac{4}{5}$

Substitute into [1]: . $p\left(P\,|\,\overline{Q}\right) \;=\; \frac{\frac{3}{20}}{\frac{4}{5}} \;=\;{\color{blue}\frac{3}{16}}$

Quote:

3) Find the probability that both $P$ and $Q$ are in the wrong boxes. (13/20)

This is a bit trickier . . .

There are two scenarios:
. . (1) $P$ goes to $Q.$
. . (2) $P$ does not go to $Q.$

(1) $P$ goes to $Q.$

. . .We have: . $\_\;\;P\;\_\;\;\_\;\;\_$

. . The other four items can have any order
. . and $Q$ will be in the wrong box.

. . .Hence: . $p(P\to Q) \;=\;\boxed{\frac{1}{5}}$

(2) $P$ does not go to $Q.$

. . . . . . . . . . . . $\overbrace{\qquad}^P$
. . . $\text{We have: }\;\;\_\;\;\_\;\;\_\;\;\_\;\;\_$

. . . $P$ can go to any of the last three addresses: . $\frac{3}{5}$
. . .Then $Q$ can go to three of the remaining four addresses: . $\frac{3}{4}$
. . . Hence: . $p(P \not\to Q) \;=\;\frac{3}{5}\cdot\frac{3}{4} \;=\;\boxed{\frac{9}{20}}$

Therefore: . $p\left(\overline{P}\,\wedge\,\overline{Q}\right) \;=\;\frac{1}{5} + \frac{9}{20} \;=\;{\color{blue}\frac{13}{20}}$
May 17th 2010, 05:01 PM
kingman

thanks

Dear Sir,

Thanks very much for the neat and clear answer. However my main problem is how to express the outcomes,events and probabilities in the Venn diagrams or tree diagram. I wonder how I can express the solution in the form of P(Q|P),using intersection and union symbols.
Thank you very much.
Kingman