probability

[jzw]

A box contains 100 balls, numbered from 1 to 100. If three balls are selected @ random and without replacement from the boc, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?

(a) 1/4
(b) 3/8
(c) 1/2
(d) 5/8
(e) 3/4

Ok. So 50 odd and 50 even. so the probability of selecting an odd or even number are both 1/2.

In order for the sum of the three numbers to be odd...

ODD + ODD + ODD = ODD which is 1/2 * 1/2 * 1/2 * 1/2 = 1/8
ODD + ODD + EVEN = EVEN which is 1/2 * 1/2 * 1/2 * 1/2 = 1/8
ODD + EVEN + EVEN = ODD which is 1/2 * 1/2 * 1/2 * 1/2 = 1/8
EVEN + EVEN + EVEN = EVEN which is 1/2 * 1/2 * 1/2 * 1/2 = 1/8

Now, @ this point you're supposed to add these together, and get 4/8 which is 1/2, I presume because you could say, "well either the first thing will happen, OR the second, OR the third, OR the fourth." And with an OR statement we add.

Great. My question is as follows: is there a shortcut for this type of probability question that I'm unaware of or do u just have to do it out?

If the question stem been the same with the exception "if two balls" instead of three, would the probability been 1/3?

[seal4913]

If u have three odds in a row without replacing its not. 5 x .5 x .5

Its. 5 x 49/99 x 48/98 plus .5 x 50/99 x 49/98

Wouldn't that make the answer a?

[seal4913]

The org question states with replacement which would make .5 correct

[Whitney Garner]

Hi jzw -

A couple of issues with your explanation:

Ok. So 50 odd and 50 even. so the probability of selecting an odd or even number are both 1/2.

That is TRUE for the first ball chosen, but this isn't the case for the second and the 3rd, because we have WITHOUT replacement.

In order for the sum of the three numbers to be odd...

ODD + ODD + ODD = ODD which is 1/2 * 1/2 * 1/2 * 1/2 = 1/8
ODD + ODD + EVEN = EVEN which is 1/2 * 1/2 * 1/2 * 1/2 = 1/8
ODD + EVEN + EVEN = ODD which is 1/2 * 1/2 * 1/2 * 1/2 = 1/8
EVEN + EVEN + EVEN = EVEN which is 1/2 * 1/2 * 1/2 * 1/2 = 1/8

Now, @ this point you're supposed to add these together, and get 4/8 which is 1/2, I presume because you could say, "well either the first thing will happen, OR the second, OR the third, OR the fourth." And with an OR statement we add.

If we only want the cases that give us ODD, then we don't add ALL of the choices you listed. In fact, only 2 of the cases you have listed will work:

ODD + ODD + ODD = ODD
ODD + EVEN + EVEN = ODD

Let's start with the easier case:

(ALL ODDS)
ODD + ODD + ODD = ODD
50/100 * 49/99 * 48/98 = 1/2 * 49/99 * 24/49 = 12/99

(ONE ODD, 2 EVENS)
ODD + EVEN + EVEN = ODD
50/100 * 50/99 * 49/98 = 1/2 * 50/99 * 1/2 = 25/(2*99)

**BUT wait a moment, that is assuming you pulled the odd ball first. You could have pulled it second OR third. The probabilities won't change (I'll let you prove that to yourself), but we have to add to additional options.

EVEN + ODD + EVEN = 25/(2*99)
EVEN + EVEN + ODD = 25/(2*99)


And now we're left to add them all up:

12/99 + 3*(25/(2*99))

24/198 + 75/198 = 99/198 = 1/2

** If you're not convinced on the order, go back to your list of the 4 possible outcomes: all odd, 1 even/2 odd, 2 even/1 odd, all even. If you were to take the probabilities of every outcome, then it SHOULD = 1, not 1/2!

The reason is because of the variety of ways we could pull these:
O O O
O O E
O E O
E O O
E E O
E O E
O E E
E E E

The probabilities for these would be {12/99, 25/198, 25/198, 25/198, 25/198, 25/198, 25/198, 12/99}, find the common denominators and you will see that 24 + 6*25 + 24 = 198, so THIS gives us 1 (certainty).

Great. My question is as follows: is there a shortcut for this type of probability question that I'm unaware of or do u just have to do it out?

There sure is - think about the problem this way - I'm going to take any 3 balls I pick and add them up. What are the options for the sum: odd and even. Well, does one happen more frequently than the other, let's see...

o + o + o = odd
o + o + e = even (3 cases of this depending on order)
o + e + e = odd (3 cases of this depending on order)
e + e + e = even

So, it looks like there are 8 possible outcomes, and 4 will give me an odd, the other 4 will give me an even...so half the time I'll get an even sum, 1/2 the time odd...the probability is just 1/2!

Remember that "without replacement" and "order doesn't matter" puts a monkey wrench in probability problems but we can still use logic to work it out!

Hope this helps!

Whit

[seal4913]

Thank you for clearing up my misunderstanding. I naturally assumed that the odds would be different when the numbers were replace and when they aren't.