Hello BTG
Would appreciate a little help on the question attached:
thanks in advance
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Arithmetic - probability
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800_or_bust
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Four possible cases:lucas211 wrote:Hello BTG
Would appreciate a little help on the question attached:
thanks in advance
Case 1: 3 Evens, 0 Odds - sum must be even
Case 2: 2 Evens, 1 Odd - sum must be odd
Case 3: 1 Even, 2 Odds - sum must be even
Case 4: 3 Odds, O Evens - sum must be odd
Since there are an equal number of even and odd balls, the probability of Case 1 OR Case 3 occurring must be equal to the probability of Case 2 OR Case 4 occurring. Hence, the probability is just 1/2.
As an aside, if you're interested in the exact probability of each case.
The probability of Case 1 is (1/2)^3 x 3C3 = 1/8 x 1 = 1/8.
The probability of Case 2 is (1/2)^3 x 3C2 = 1/8 x 3 = 3/8.
The probability of Case 3 is (1/2)^3 x 3C1 = 1/8 x 3 = 3/8.
The probability of Case 4 is (1/2)^3 x 3C0 = 1/8 x 1 = 1/8.
Case 1 + Case 3 = 1/8 + 3/8 = 1/2.
Case 2 + Case 4 = 3/8 + 1/8 = 1/2.
800 or bust!
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Hi lucas211,
These types of questions can be solved in a couple of different ways, depending on how you like to organize your information and how you "see" the question.
Since half the numbered balls are 'odd' and half are 'even', and we're replacing each ball after selecting it, on any given selection we have a 50/50 chance of getting an odd or even. This means that there are....
(2)(2)(2) = 8 possible arrangements when selecting 3 balls from the box:
EEE
EEO
EOE
OEE
OOO
OOE
OEO
EOO
Since we have the same number of Odds and Evens, each of these options has the same 1/8 probability of happening. Now you just have to find the ones that give us the ODD SUM that we're asked for. They are:
EEO
EOE
OEE
OOO
Four of the eight options. 4/8 = 1/2
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
These types of questions can be solved in a couple of different ways, depending on how you like to organize your information and how you "see" the question.
Since half the numbered balls are 'odd' and half are 'even', and we're replacing each ball after selecting it, on any given selection we have a 50/50 chance of getting an odd or even. This means that there are....
(2)(2)(2) = 8 possible arrangements when selecting 3 balls from the box:
EEE
EEO
EOE
OEE
OOO
OOE
OEO
EOO
Since we have the same number of Odds and Evens, each of these options has the same 1/8 probability of happening. Now you just have to find the ones that give us the ODD SUM that we're asked for. They are:
EEO
EOE
OEE
OOO
Four of the eight options. 4/8 = 1/2
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
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OptimusPrep
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Given: 100 balls. 3 balls are selected with replacement from the box.lucas211 wrote:Hello BTG
Would appreciate a little help on the question attached:
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
thanks in advance
Required: Sum of the balls has to be odd.
The sum will be odd if:
3balls odd
2 even 1 odd
Total cases: 3odd, 1even 2 odd, 2 even 1 odd, 3 odd = 4
Probability that the sum will be odd = 2/4 = 1/2
Correct Option: C
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Matt@VeritasPrep
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p(all odd) = p(all even)
p(one odd) = p(one even)
Suppose we say the first = x and the second = y. We want
p(all odd) + p(one odd)
----------------------------------------------------
p(all odd) + p(all even) + p(one odd) + p(one even)
which is just
(x + y)
--------
(x+x+y+y)
or
(x + y)/(2x + 2y)
or
1/2
p(one odd) = p(one even)
Suppose we say the first = x and the second = y. We want
p(all odd) + p(one odd)
----------------------------------------------------
p(all odd) + p(all even) + p(one odd) + p(one even)
which is just
(x + y)
--------
(x+x+y+y)
or
(x + y)/(2x + 2y)
or
1/2
