# Thread: pls help me w probability!

1. ## pls help me w probability!

1) 24 identical balls of two different colors are placed into two bags. Bag A contains 6 blue and 6 green balls and Bag B contains 3 blue and 9 green balls. One ball is taken out from each bag. Find the probability that both balls are of the

same color.

2) The library records that 90% of the borrowed books were returned on time while 7% were returned late and the rest not returned at all. 5% of the books returned on time were torn while only 2% of those returned late were torn. Find the probability thata book was returned in good condition.

3)Tom and Henry meet to have a game of chess. It is known that when Tom has the first move in the game, the probability of Tom winning the chess game is 0.4 and the probability of Henry winning that game is 0.1. When Henry has the first move in a game, the probability of Henry winning that game is 0.3 and the probability of Tom winning that game is 0.2. A draw occurs when neither of the players wins. Given that Tom and Henry toss a fair coin to decide who has the first move in the game, find the probability of the game ending in a draw.

4)The probability that it rains is 0.3. On a rainy day, the probability that the bus is late is 0.25. On a sunny day, the probability that the bus is on time is 0.95. Given that the bus arrives on time, find the probability that it is a rainy day.

2. Hi

1)

A="Both balls are blue"
B="Both balls are green"

$P(A)= \left(\frac{6}{12}\right) \cdot \left(\frac{3}{12}\right)$

$P(B)=\left(\frac{6}{12}\right) \cdot \left(\frac{9}{12}\right)$

$P(A)+P(B) = 0.5$

2)

I calculated the probability that a book was torn, and took 1 minus that probability.

$1-(0.9\cdot 0.05 + 0.07 \cdot 0.02) = 0.9536$

3) Since the coin is fair, it is a 50-50 shot that either of them will go first. But given the probabilites here it doesnt matter who goes first, it is always 50-50 for a draw. So...

Answer: $0.5$

4)

A="The bus is not late"
B="It does not rain"

Then by Bayes Theorem: $P(B|A)=\frac{P(A|B)\cdot P(B)}{P(A)}=\frac{0.95\cdot 0.7}{1-(0.3\cdot 0.25+0.7\cdot 0.05)} = 0.747$

Here we calculated "Given that the bus was not late, what is the probability of it not raining". So therefore we must take $1-0.747=0.253$ to get the probability of it raining.

I hope these are right, I´m confident mrFantastic or anyone more experienced will soon give their input as well.

3. Thanks so much.
But is there other method beside using Bayes Theorem for Q4?
Because I've never learn that theorem before, therefore i couldn't understand your working.
But the answer is correct by the way.

4. Hi

I really don´t if there are any other ways, if there are, I don´t know them I'm afraid.

But you should really look into Bayes theorem:

Bayes' theorem - Wikipedia, the free encyclopedia

There is a good example there as well.

GL

5. Hello, dorwei92!

I have a different answer for #2 . . .

2) The library records that 90% of the borrowed books were returned on time
. . while 7% were returned late and the rest not returned at all.
5% of the books returned on time were torn, while only 2% of those returned late were torn.

Find the probability that a book was returned in good condition.
We have these probabilities:

. . $\begin{array}{c||c|c|c|} \hline \\[-4mm]
\text{On time} & P(OT) \:=\:0.90 & P(\text{ torn}) \:=\:0.05 & P(\text{good}) \:=\:0.95 \\ \hline \\[-4mm]
\text{Late} & P(\text{Late}) \:=\:0.07 & P(\text{ torn}) \:=\:0.02 & P( \text{good}) \:=\:0.98 \\ \hline\end{array}$

We have: . $\begin{array}{ccccc}P(OT\,\wedge\, \text{good}) &=& (0.90)(0.95) &=& 0.8550 \\ \\[-4mm]
P(\text{Late}\,\wedge\,\text{good}) &=& (0.07)(0.98) &=& 0.0686\end{array}$

Therefore: . $P(\text{good}) \;=\;0.8550 + 0.686 \;=\;0.9236$

6. Hi

Soroban, I am so stupid.

I forgot about the books that does not return at all!

7. I've come to have a little understanding about Bayes' Theroem through some videos on youtube.
But I still does not really understand why must it it
P(B/A) and not P(A/B)? Isn't the both of them same?
Could u explain a little further for me?

And also the Bayes theorem that u did was to calculate the probability of it not raining and bus was on time...so that later on you take 1 to minus the result from the calculation to get to the solution; bus was not late and it's a rainy day?

8. Originally Posted by dorwei92
I've come to have a little understanding about Bayes' Theroem through some videos on youtube.
But I still does not really understand why must it it
P(B/A) and not P(A/B)? Isn't the both of them same?
Could u explain a little further for me?

[snip]
Do you think Pr(King | Royal card) is the same as Pr(Royal card | King) ....??

(How much conditional probability have you learned?)

9. Both of you were really of great help to me.
Thanks so much!!!
And Twig you ain't stupid at all, human beings are bound to make mistakes at times...but im really grateful anyway!!
THANK YOU!

10. Originally Posted by mr fantastic
Do you think Pr(King | Royal card) is the same as Pr(Royal card | King) ....??

(How much conditional probability have you learned?)
Honestly, I have not learn conditional probability and bayes theorem before at all, so there is a limit to my understanding about this topic.
I seriously don't get it...im so sorry.
Could u explain for me then...thanks.

11. Hi

An example on conditional probability:

Say you have a deck of cards. Now, if you draw one card, the probability of it being an ace is $\frac{4}{52}$ , since there are 4 aces in a deck of total 52 cards.

Say you drew one card and you got the ace of spades. Now you decide to draw another card. What is the probability of the second card being an ace?

It cannot be $\frac{4}{52}$ obviously, since there are only 3 aces left, and also, the deck only consists of a total of 51 cards now.

So, the probability of the second card being an ace is $\frac{3}{51}$ . Here it is IMPORTANT to realize that this is a conditional probability.

Because if the first card you drew was not an ace, then there would be a bigger chance of the second card being one, right?

So, to summarize:

A="The first card is an ace"
B="The second card is an ace"
C="The first card is not an ace"

$P(B|A)=\frac{3}{51}$ , read: "B given A"

$P(B|C)=\frac{4}{51}$ , because here you have 4 aces left in a deck of 51 cards, because the first card was not an ace!

12. I finally understand Bayes' theorem through videos and wiki explanations.
thanks Twig you "indirectly" increases my knowledge capacity...
Now that i learned about this theorem, im sure i would be able to tackle such questions easily in the future!~