1. probability

1) A school offers japanese and arabic as two foreign languages. the probabilitiy that a student who studies japanese also studies arabic is 1/4 and 1/5 of the students who study arabic also studies japanese too. among the students, 1/5 of them do not study any of the two languages . what is the probability that a student in the school studies both languages? Are the events of studying japanese and arabic independent? ( answer provided 1/10,not independent) explain pls

2)Bag A contains 3 red balls and 2 green balls whereas bag B contain 2 red balls and 3 green balls. two balls r withdrwan from bag a to be put into bag b . then two balls are taken out from B to be put in bag A. what is the probability that bag A now contains 3 red balls and 2 green balls.? answer 10/21

2. Originally Posted by qweiop90
1) A school offers japanese and arabic as two foreign languages. the probabilitiy that a student who studies japanese also studies arabic is 1/4 and 1/5 of the students who study arabic also studies japanese too. among the students, 1/5 of them do not study any of the two languages . what is the probability that a student in the school studies both languages? Are the events of studying japanese and arabic independent? ( answer provided 1/10,not independent) explain pls

[snip]
Draw a Karnaugh table:

$\begin{tabular}{l | c | c | c} & J & J\, ' & \\ \hline A & a & b & \\ \hline A\, ' & c & 1/5 & \\ \hline & & & 1 \\ \end{tabular}
$

From the given data you know that:

$\frac{a}{a+b} = \frac{1}{5} \Rightarrow b = 4a$ .... (1)

$\frac{a}{a+c} = \frac{1}{4} \Rightarrow c = 3a$ .... (2)

$a + b + c = \frac{4}{5}$.

Solve for a, b and c.

From the Karnaugh table $\Pr (A \cap J) = a$.

A and J are independent if $\Pr (A \cap J) = \Pr(J) \cdot \Pr(A)$.

$\Pr(J) = a + c = \, ....$
$\Pr(A) = a + b = ....$

Does $\Pr (A \cap J) = \Pr(J) \cdot \Pr(A)$ ....?

3. Originally Posted by qweiop90
[snip]
2)Bag A contains 3 red balls and 2 green balls whereas bag B contain 2 red balls and 3 green balls. two balls r withdrwan from bag a to be put into bag b . then two balls are taken out from B to be put in bag A. what is the probability that bag A now contains 3 red balls and 2 green balls.? answer 10/21
Two balls taken from bag A. The possibilities are:

2R => Bag B has 4R and 3G. Probability = (3/5)(2/4) = 6/20. Bag A has 1R, 2G.

1R, 1G => Bag B has 3R and 4G. Probability = 2 (3/5)(2/4) = 12/20. Bag A has 2R, 1G.

2G => Bag B has 2R and 5G. Probability = (2/5)(1/4) = 2/20. Bag A has 3R.

Now consider each of the possible bag B's:

4R, 3G: The two balls taken need to be 2R for Bag A to have 3R and 2G. The probability of this is (4/7)(3/6) = 12/42.

3R, 4G: The two balls taken need to be 1R 1G for Bag A to have 3R and 2G. The probability of this is 2 (3/7)(4/6) = 24/42.

2R 5G: The two balls taken need to be 2G for Bag A to have 3R and 2G. The probability of this is (5/7)(4/6) = 20/42.

So the total probability is (6/20)(12/42) + (12/20)(24/42) + (2/20)(20/42) = .......