Math Is Fun Forum

qweiop90

Member

Registered: 2008-08-19

Posts: 8

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

gnitsuk

Member

Registered: 2006-02-09

Posts: 121

Re: probability

Question (2)

Okay, so we want the probability of restoring bag A to its original contents after randomly taking two from bag A and placing them in B and then randomly taking two from B and placing them in A.

We could do this in only three ways:

Event A: Take two red from bag A and put them in bag B and then take two red from bag B and put the in bag A

OR

Event B: Take two green from bag A and put them in bag B and then take two green from bag B and put the in bag A

OR

Event C: Take one red and one green from bag A and put them in bag B and then take one red and one green from bag B and put them in bag A

So our answer is:

P(Restoring Bag A) = P(Event A) + P(Event B) + P(Event C)

Because we can restore bag A only by acheiving Event A OR Event B OR Event C.

Okay,

Event A:

Probability of taking a red from bag A = 3/5 and so remaining probability of taking a red = 2/4 and so probability of taking two reds is (3/5)*(2/4) = 3/10

Now bag B contains four reds and three greens. So probability of taking a red from bag B = 4/7 and so remaining probability of taking a red from bag B is 3/6 and so probability of taking two reds from B is 4/7 * 3/6 = 2/7

And so finally, probability of restoring bag A in this way is the probability of BOTH these things happening = 3/10 * 2/7 = 3/35

Event B:

This is calculated in exactly the same way but using green and gives 1/21

Event C:

Here, the probability of taking one red and one green from bag A is:

3/5 * 2/4 + 2/5 * 3/4  = 3/5 becasue we might take one red then one green OR one green then one red

These are put into B and in a similar way probability of taking one green and one red from B is now 3/7*4/6 + 4/7 * 3/6 = 4/7

Multiply these to give probability of restoring A in this way = 3/5 * 4/7 = 12/35

Finally add all the results to give:

3/35 + 1/21 + 12/35 = 10/21

Last edited by gnitsuk (2008-08-20 04:26:46)

gnitsuk

Member

Registered: 2006-02-09

Posts: 121

Re: probability

Question (1)

Here, we need to use the conditional probability formula which states that:

The probability  of Event A occuring given that Event B has occured is equal to the probability of events A and B occuring divided by the probability of event B, so in symbols this is written:

Okay, so lets label our terms:

Let Probability of a student studying Arabic = P(A) = a
Let Probability of a student studying Japanese = P(J) = j
Let Probability of a student studying both Arabic and Japanese = P(AnJ) = b

We want to find b.

Well, we are told the following facts:

Note here that in the above formulae we use the obvious fact that P(AnJ) = P(JnA)

In this last equation we see that the probability of studying no language is one less the probability of studying Arabic only less probability of studying Japanese only plus probabiilty of studying both (a Venn diagram will make this clear to you).

So swapping to our a,b,c notation this gives:

Three equations in three unknowns. Solving these simultaneously gives the desired:

The events of studying Japanese and studying Arabic or not independent, a student may be studying either or both.

Last edited by gnitsuk (2010-05-28 02:17:45)