Probability

Prakash Panneer · 2006-12-20 11:07:12

1. An urn contains 40 balls numbered 1 through 40. Suppose that numbers 1 through 10 are considered "lucky." Two balls are drawn from the urn without replacement. Find the probability that
(a) both balls drawn are "lucky";
(b) neither ball drawn is "lucky";
(c) at least one of the balls drawn is "lucky";
(d)exactly one of the balls drawn is "lucky."

2.If a man having a batting average of 0.40 comes to bat 5 times in a game, what is the probability that he will get
(a) exactly 2 hits;
(b) less than 2 hits?

Thanks in advance

John E. Franklin · 2006-12-20 12:10:11

1a) I did all the cases from 3 to 5 balls, picking two on paper and learned a few things.
Then I found the probability for 1a) is 45 / [ (19)(40)+20 ], which is also equal to the simpler (1/4)(9/39).
Each are done different ways. The first thru drawing out many things, and the second is just guessing the formula is first 10 out of 40, then 9 out of 39 multiplied together. It appears they are both the same.

Also my first way is simply all the numbers from 1 to 9 added up divided by the sum of the numbers from 1 to 39.

Now for 1b), it should be the sum of all numbers from 1 to 29 divided by the sum of all numbers from 1 to 39.
At least that's what I get by drawing combinations on paper and keep adding a new person to the list.
I use people I know instead of numbers, it's easier to keep track of.

Last edited by John E. Franklin (2006-12-20 12:20:43)

John E. Franklin · 2006-12-20 14:31:43

1d) This should be 100% minus 1a and minus 1b. Because if there are 3 possibities, and 1 of each lucky or not is this third possiblity.

1c) This should be 100% minus 1b answer.

I hope this is right, or else I'm wasting your time, huh?

By the way, my sketchy notes on the combinations are:
PM MP

PJ JP
MJ JM

PB BP
MB BM
JB BJ

AB BA
AJ JA
AM MA
AP PA

There are 5 people P, M, J, B, and A, added to the groups in that order.
The pairs are two picked from them.

John E. Franklin · 2006-12-20 14:40:47

2a) I want to dispute this question because if the guy has a lucky bat for 40% of his games he plays, and for those games, he hits every time, then the distribution is such that the 400 batting average is not random.
And hence the exactly 2 hits would never happen, it would depend if he held the lucky bat!!!
Gotcha!!!! (chuckle)

Ricky · 2006-12-20 14:55:36

1. An urn contains 40 balls numbered 1 through 40. Suppose that numbers 1 through 10 are considered "lucky." Two balls are drawn from the urn without replacement. Find the probability that
(a) both balls drawn are "lucky";
(b) neither ball drawn is "lucky";
(c) at least one of the balls drawn is "lucky";
(d)exactly one of the balls drawn is "lucky."
2.If a man having a batting average of 0.40 comes to bat 5 times in a game, what is the probability that he will get
(a) exactly 2 hits;
(b) less than 2 hits?

1a. Drawing the first ball has a 1/4 chance to be lucky (10/40). Note that if we don't draw the first ball lucky we fail no matter what, and thus, we don't have to try to count the case that the first ball was unlucky. Under the assumption that the first ball was lucky, we now have a 9/39 chance to draw another lucky ball. This ends up being 10/40 * 9/39 since these events are independent.

1b. Do the same as the above. Just replace the chances for lucky with unlucky. Note that P(lucky) + P(unlucky) = 1

1c. Ohhh, now things get interesting. We have four cases:

1st and 2nd lucky
1st lucky, 2nd unlucky
1st unlucky, 2nd lucky
1st and 2nd unlucky

Note that the last one does not satisfy our condition of drawing at least one lucky ball, so we disgaurd it. Now just as above, calculate each one of the above (all 3). Each of these are mutually exclusive, and thus, add them together and you will get the overall chance.

1d. Do the same as 1c, but you now also disgaurd the case where both the 1st and 2nd are lucky.

For 2, do you know binomial probabilities? If so, I recommend using it. I'm not quite sure how to approach such a problem without using it. If you need help using binomial probability, just let me know.

John E. Franklin · 2006-12-20 14:57:24

Okay, I'm ready to work on this #2 problem, but I prefer to rewrite it like this:
A thousand sided die is marked with 400 of its sides labelled "HIT", and 600 of
its sides labelled "NOHIT".
The die is rolled five times. What are the possible outcomes in terms of "HIT" and "MISS" given along with the percentage chance information. Iterate all 32 possibilities, time permitting.
(2 up 5)

John E. Franklin · 2006-12-20 15:04:46

Hint: Relabel the sides with 400 called "(.4)" and 600 called "(.6)".
The point 4 and point 6 will be used in the math later on, probably multiplied as I iterate the 32 possiblities and
group them to find how many are 5 choose 2, since I never memorized that formula yet, I have to invent it.

John E. Franklin · 2006-12-20 15:34:18

2a)

The 4 means a hit and represents .4 times the digits beside it. The 6 means .6 or no hit.
Here are the ten permutations with two fours out of 32 permutations.
44666 = 0.03456
46466 = 0.03456, just multiply .4 x .6 x .4 x .6 x .6
46646 = 0.03456
46664
64466
64646
64664
66446
66464
66644
That's ten of them, so ten times .03456 is 0.3456 probability!!!!!!!!
So 2a) is 34.56% chance of exactly 2 Hits out of 5.
At least I hope so!!!!

John E. Franklin · 2006-12-20 15:48:10

2b) Less than 2 hits answer:
46666 1 hit followed by 4 misses, probability = .4 .6 .6 .6 .6 = .05184
64666 1 miiss, a hit, and then 3 misses = .6 x .4 x .6 x .6 x .6 = 0.05184
66466 1 hit in middle of five up at bats = again 0.05184
66646 1 hit on 4th at bat is again 0.05184
66664 1 hit on last 5th at bat is again 0.05184
66666 (no hits) Chance of no hits is .6 up 5 or 60% to the power of 5.
This is .07776 for 0.6^5.
So sum them all up and you get 5 times .05184 plus .07776 for the answer of:
0.33696 Hopefully!!!!!!!! (for under 2 hits, Chance may be 33.696%)

pi man · 2006-12-20 16:50:06

I'll use a different approach. How many different sets of 2 balls could you possibly choose? That would be (40 choose 2) ways or 780 ((40 * 39) / (2*1)).

How many different ways involve 2 lucky draws? That would be (10 choose 2) or 45.
How many different ways involve 2 unluck draws? That is (30 choose 2) or 435.
How many for 1 lucky and 1 unlucky? (10 choose 1) * (30 choose 1) or 300.

45 + 435 + 300 = 780. That's a good thing.

So:
a) both lucky: 45/780 = .057692
b) both unlucky: 435 / 780 = .55769
d) one lucky and one unlucky: 300 / 780 = .38461
c) one or more lucky: (300 + 45) / 780 = .44231

Math Is Fun Forum

#1 2006-12-20 11:07:12

Probability

#2 2006-12-20 12:10:11

Re: Probability

#3 2006-12-20 14:31:43

Re: Probability

#4 2006-12-20 14:40:47

Re: Probability

#5 2006-12-20 14:55:36

Re: Probability

#6 2006-12-20 14:57:24

Re: Probability

#7 2006-12-20 15:04:46

Re: Probability

#8 2006-12-20 15:34:18

Re: Probability

#9 2006-12-20 15:48:10

Re: Probability

#10 2006-12-20 16:50:06

Re: Probability

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