Question about Texas Instruments BA-II Plus Calculator

# What is the Expexted Return ? Probability .29 Return 39% Probability .55 Return 22 Probability .16 Return -20

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( . 2 9 * . 3 9 ) + ( . 5 5 * . 2 2 ) + ( . 1 6 * . 2 +/- ) =

In AOS mode you can skip the parentheses.

Posted on Feb 26, 2011

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### What are the steps in solving for probability on C?

If you have nine marbles in a bag, four of which are blue, then the probability of reaching in and pulling out a blue marble is four out of nine, or 4 divided by 9.
Stay in the normal computational mode and press 4 / 9 =
If you want to know the probability of pulling out three (for example) marbles and getting at least one blue marble, then you get into combinatorics. The answer depends on whether or not you replace each marble after pulling it out (Sampling with/without replacement).
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### How many people must be in a room before the probability that some share a birthday becomes at least 50 percent? Use different values of n in the formula and approximate to the nearest n. show all work.

With two people in the room, the probability that the second person shares a birthday with the first is 1/365 (let's ignore leap years for the time being). Thus, the probability that they DON'T share a birthday is 364/365.

Add a third person. The probability that this person does not share a birthday with either of the first two (assuming they don't share a birthday) is 363/365. Thus, the probability that none of the three share a birthday is (364/365)*(363/365).

The probability that a fourth person doesn't share a birthday with any of the first three is 362/365. Thus the probability that none of the four share a birthday is (364/365)*(363/365)*(362/365).

See the pattern here? The probability of the n-th person not sharing a birthday is (366-n)/365. Keep adding people and multiplying the terms together until the probability of no one sharing a birthday drops below 0.5 --- which is when the probability of at least one pair sharing a birthday climbs above 50%.

Since this sounds like a homework problem, you'll have to do the rest.

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### Getting Weighted Std Dev for probability..

Sorry, boss.

But getting standard deviation from a probability distribution would probably be best done by hand (of course, with the use of a computational device for those hard-to-reach fractions).

If your distribution question has a long line of events (IE, data points) then you'd probably want to use a more advanced calculator.

Nevertheless....I will return this afternoon with a more informed answer after I tinker with my BA-II Plus.

Whenever I do these, though, I usually pop open my TI89 Titanium (or of course just do it by hand).

STAY TUNED, MO...

Mar 11, 2011 | Texas Instruments BA-II Plus Calculator

### Expected returns: You have chosen biology as your college major because you would like to be a medical doctor. However, you find that the probability of being accepted to medical school is about 10...

So expected returns are just a little more advanced version of probability - without getting too in detail it uses weighted averages so you do not have to add the same number 17 times.
Let's first define our variables:
E(X) = Expected Return X1 = Variable 1 P1 = Probability 1 (in decimal format so 55% would be .55) X2 = Variable 2 P2 = Probability 2 (P2 is also equal to 1 - P1)
So the expected return formula is as follows:
E(X) = (X1)*(P1) + (X2)*(P2)
Plugging in our given numbers we get E(X) = Expected Return X1 = 300,000 P1 = 10% or .1 X2 = 40,000 P2 = 1 - P1 = 1 - .1 = .9
E(X) = (300,000)*(.1) + (40,000)*(.9) E(X) = 66,000
Now remember this does not include the opportunity cost of spending your four years in college studying biology instead of another degree.

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### How do i generate random integers with a ti 89 titanium?

Use the rand() function. If you give it a positive integer argument n, it will return a random integer in the range in the interval [1, n]. You'll find rand() in the MATH/PROBABILITY menu.

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### A university has to select an examiner from a list of 50 persons, 20 of them are women and 30 men, 10 of them knowing Hindi and 40 not, 15 of them being teachers and remaining 35 not. What is the...

To get probability of one result and another from two separate experiments, multiply the individual probabilities.

Probability of finding a woman = 20/50

Probability of finding one knowing Hindi = 10/50

Probability of finding a teacher = 15/50

Therefore, probability of the university selecting a Hindi knowing women teacher =

20/50 x 10/50 x 15/50 = 3/125 or 0.024 or about 1 in 50 (1.2/50)

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### I have three tenders a, b and c estimate probability it secure tender a is 0.2 b is 0.3 and c is 0.4 what the probability of win only one tender

There can be three possibility for winning only one tender. Can win only a or only win b or only win c.

probability for winning a = 0.2.

probability of loosing a = 1 - 0.2 =0.8.

Similarly probability of loosing b and c are 0.7 and 0.6 respectively.

so probability of winning only one tender = win a * lose b * lose c + lose a* win b * lose c + lose a * lose b * lose c.

= 0.2 * 0.7 * 0.6+ 0.8 * 0.3 * 0.6 + 0.8 * 0.7 * 0.4

= .452 is the probability of getting only one tender.

Thank you.

Jun 11, 2010 | Mathsoft Solving and Optimization...

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