Problem Solving

A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, 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


I wanted to see how others approached this problem. Luckily I've posted and practiced many many probability questions in this forum, so I got this one right. However I did it by inspection, so if others could answer it I can see if my approach is good in general or simply worked for this particular question.

Ok number of ways the sum of 3 digits can be odd:

E+E+O
O+O+O

(E - Even, O - Odd)

E E O can further be arranged in 3 ways.

So total probability = (1/2*1/2*1/2)*3 + (1/2*1/2*1/2) = 3/8+1/8 = 4/8 = 1/2.

IMO C.

there are 50 odd balls and 50 even balls

and the possible results are either odd or even

possible combination for ODD result :

O + E + E & O + O + O

possible combination for even result :

E + O + O & E + E + E

hence the probability would be 1/2

imo, answer C .

what is OA?

Here is the approach I took:

Here are the possible outcomes for the balls:

E+E+E = Even
E+E+O = Odd
E+O+O = Even
E+O+E = Odd
O+E+E = Odd
O+E+O = Even
O+O+E = Even
O+O+O = Odd


4 possibilities it will be even and 4 that it will be odd

so 4/8 that it will be odd

1/2 that it will be odd

Alternatively, _ _ _ (need to fill w/ E or O)

this is like a coin flip question, so total ways = 2^n = 2^3 = 8
Half of these will add to even and the other half to odd, so probability = 1/2

3 numbers to choose and 2 choices(EVEN OR ODD) for each number

2*2*2 = 8 different outcomes

OOE
OEO
EOO
EEE
OOO
EEO
EOE
EOO

3/8 cases ithe sum will be odd

cramya wrote:3 numbers to choose and 2 choices(EVEN OR ODD) for each number

2*2*2 = 8 different outcomes

OOE
OEO
EOO
EEE
OOO
EEO
EOE
EOO

3/8 cases ithe sum will be odd

OOE=even
OEO=even
EOO =even
EEE =even
OOO =odd
EEO =odd
EOE =odd
EOO ( should be OEE not EOO so this would be odd)

The OA is 1/2.

Everyone has the right idea. We know that the probability of even or odd is 1/2.

dmateer25 wrote:Here is the approach I took:

Here are the possible outcomes for the balls:

E+E+E = Even
E+E+O = Odd
E+O+O = Even
E+O+E = Odd
O+E+E = Odd
O+E+O = Even
O+O+E = Even
O+O+O = Odd

The OA is 1/2.

Since the order doesn't matter (with replacement), you can see here that 1/2 of your options will give you even or odd. If the question was without replacement then you would need to do some calculations. I guess either way you get the right answer by doing it like a coin flip but inspection, if you can do it, saves time.

What would be the answer w/o replacement?

I'm confused by what you mean when you say "order doesn't matter." If order doesn't matter then E-E-O and O-E-E are the same thing. But then why are we listing them out to be different possible outcomes?

evansbd wrote:The OA is 1/2.

Everyone has the right idea. We know that the probability of even or odd is 1/2.

dmateer25 wrote:Here is the approach I took:

Here are the possible outcomes for the balls:

E+E+E = Even
E+E+O = Odd
E+O+O = Even
E+O+E = Odd
O+E+E = Odd
O+E+O = Even
O+O+E = Even
O+O+O = Odd

The OA is 1/2.

Since the order doesn't matter (with replacement), you can see here that 1/2 of your options will give you even or odd. If the question was without replacement then you would need to do some calculations. I guess either way you get the right answer by doing it like a coin flip but inspection, if you can do it, saves time.

What would be the answer w/o replacement?

evansbd wrote:A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, 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


I wanted to see how others approached this problem. Luckily I've posted and practiced many many probability questions in this forum, so I got this one right. However I did it by inspection, so if others could answer it I can see if my approach is good in general or simply worked for this particular question.

Should be C

Why can't your thinking just be half the numbers from 1-100 are odd ? Forget about sum for a minute. What's the probably of any ball that's picked will be odd right? Half the balls are odd. Half the balls are even. nuff said? I don't see the need for all of the E+O+E etc exercise.

abcdefg wrote:I'm confused by what you mean when you say "order doesn't matter." If order doesn't matter then E-E-O and O-E-E are the same thing. But then why are we listing them out to be different possible outcomes?

evansbd wrote:The OA is 1/2.

The OA is 1/2.

Since the order doesn't matter (with replacement), you can see here that 1/2 of your options will give you even or odd. If the question was without replacement then you would need to do some calculations. I guess either way you get the right answer by doing it like a coin flip but inspection, if you can do it, saves time.

I'm actually also struggling with that question: why isn't EOO treated the same way as OOE if order doesn't matter?