# Normal Distribution

• Mar 11th 2010, 03:03 AM
charikaar
Normal Distribution
Suppose that the probability that the price of a commodity goes up today is 0.6 that the probability that it goes up tomorrow is 0.7, and that the probability it goes up both today and tomorrow is 0.5. Find the probability that it goes down or remains the same both today and tomorrow.

Let A =the price goes up today and B=price goes up tomorrow.

we have P(A)=0.6, P(B)=0.7, P(AandB)=0.5

Do they want 1-P(AandB)=0.5?

Thanks
• Mar 12th 2010, 04:40 AM
crymorenoobs
Yes, if you think of the commodity price change on a number line... it going up would be (>), the only other 2 possibilities for it would be going down (<) or remaining the same (=).

So if C = the price goes down today, D = the price goes down tomorrow,
and E = the price stays the same today, F = the price stays the same tomorrow.

Then the probability that the prices go down *or* remain the same today *and* tomorrow is : P((C or E) and (D or F))

P((C or E) and (D or F)) = P(NOT(A and B))
= 1 - P(A and B)
= 1 - 0.5
= 0.5
• Mar 12th 2010, 10:49 PM
CaptainBlack
Quote:

Originally Posted by charikaar
Suppose that the probability that the price of a commodity goes up today is 0.6 that the probability that it goes up tomorrow is 0.7, and that the probability it goes up both today and tomorrow is 0.5. Find the probability that it goes down or remains the same both today and tomorrow.

Let A =the price goes up today and B=price goes up tomorrow.

we have P(A)=0.6, P(B)=0.7, P(AandB)=0.5

Do they want 1-P(AandB)=0.5?

Thanks

There are four possibilities for the price on the two days: $\displaystyle uu$, $\displaystyle ud,\ du,\ dd.$ Where $\displaystyle d$ denotes the event the the price does not go up on a particular day. Then:

$\displaystyle p(uu)+p(ud)+p(du)+p(dd)=1$

The probability that the price goes up on the first day is:

$\displaystyle p(u\cdot)=p(uu)+p(ud)$

but you are told that $\displaystyle p(u\cdot)=0.6$ and that $\displaystyle p(uu)=0.5$, so $\displaystyle p(ud)=0.1$

Similarly that the proce goes up on the second day is:

$\displaystyle p(\cdot u)=p(uu)+p(du)$

and as $\displaystyle p(\cdot u)=0.7$, $\displaystyle p(du)=0.2$

Now we are asked for:

$\displaystyle p(dd)=1-(p(uu)+p(ud)+p(du))$

which you can compute from what we now know.

CB