Probability

Aladdin

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An urn U1 contains 5 black balls and 4 white balls.
An urn U2 contains 4 black balls and 5 white balls

Part A.
We draw simultaneously 3 balls from U1.
1) Calculate the probability to get three balls of the same color.
2)Calculate the probability to get at least one black ball.

Part B.
We choose randomly one of the two urns and we draw Simultaneously two balls from the chosen urn.
Consider the following events :
E:The chosen urn is U1
F:The drawn balls are white.
1) Calculate P(F/E) and P(F).
2)The two drawn balls are white, calculate the probability that they come from U1?


My achievments so far : :
Part A.
!)P(3 balls of the same color from U1) = 5C3/9C3 + 4C3/9C3
2)Pr(at least 1 black) = all - non = 1 - 5C3/9C3

For Part B)
1)Pr(F/E) = 4C2/9C2
Pr(F) = 4C2/9C2 + 5C2/9C2.
2) P = P(F)/2.

I'm in too much need for help , I enjoy doing probability problems but I usually get them wrong !!
Notes : What's the difference between Simultaneously and successivly.

Many Thanks.
Aladdin
 
Part A (2) is correct if the balls are chosen from Urn2 but not Urn1.
 
Hello, Aladdin!

Urn \(\displaystyle U_1\) contains: 5 black balls and 4 white balls.
Urn \(\displaystyle U_2\) contains: 4 black balls and 5 white balls

Part B.
We choose randomly one of the two urns and we draw simultaneously two balls from the chosen urn.

Consider the following events :
. . E: The chosen urn is \(\displaystyle U_1\)
. . F: The drawn balls are white.

1) Calculate P(F|E) and P(F).

\(\displaystyle (1)\;P(WW\,|\,U_1) \;=\;\frac{_4C_2}{_9C_2}\) . . Correct!

. . . \(\displaystyle P(WW) \:=\:\frac{_4C_2}{_9C_2} + \frac{_5C_2}{_9C_2}\) . . Ritght!



2) The two drawn balls are white.
. . Calculate the probability that they come from U1.

\(\displaystyle P(U_1\,|\,WW) \;=\;\frac{P(U_1\,\wedge\,WW)}{P(WW)}\)


\(\displaystyle \text{The numerator is: }\:p(U_1 \wedge WW) \:=\:\frac{1}{2}\!\cdot\!\frac{_4C_2}{_9C_2} \;=\;\frac{1}{12}\)

\(\displaystyle \text{The denominator is: }\:p(WW) \;=\;\frac{1}{2}\!\cdot\!\frac{_4C_2}{_9C_2} + \frac{1}{2}\!\cdot\!\frac{_5C_2}{_9C_2} \;=\;\frac{2}{9}\)


\(\displaystyle \text{Therefore: }\;P(U_1\,|\,WW) \;=\;\frac{\frac{1}{12}}{\frac{2}{9}} \;=\;\frac{3}{8}\)


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If a number of balls is drawn simultaneously,
. . they are drawn at the same time.
There is no opportunity to replace any of the drawn balls.


If a number of balls is drawn successively,
. . they are drawn one at a time.
The problem must clearly state if the balls are replaced after each draw
. . or if they are drawn without replacement.

 
Thanks for the help Soraban and Cris . . .

I finished my Exams doing almost great.So I want to say thank you for all members here on Free Math help forum for their advice and wisdom they shared with me.

I'm seeking for a college that suits my advantages . . .
 
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