Quote:
Harsh2111s wrote:
Quote:
Original question is:
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
The sum of the three numbers on the balls selected from the box to be odd one should select either three odd numbered balls (Odd+Odd+Odd=Odd) or two even numbered balls and one odd numbered ball (Even+Even+Odd=Odd);
P(OOO)=(1/2)^3;
P(EEO)=3*(1/2)^2*1/2=3/8 (you should multiply by 3 as the scenario of two even numbered balls and one odd numbered ball can occur in 3 different ways: EEO, EOE, or OEE);
So finally P=1/8+3/8=1/2.
Answer: C.
Alternately you can notice that since there are equal chances to get even or odd sum after two selections (for even sum it's EE or OO and for odd sum it's EO or OE) then there is 1/2 chances the third ball to make the sum odd.
Bunuel,
Need some help here. The question clearly states
what is the probability that
the sum of the three numbers on the balls selected from the box will be odd ?
Now Either I choose EEO or OEE or EOE, in each case sum of three numbers is ODD.
Then Why should I consider the order in which ball is taken out ?
2even+1odd=odd
3odd=odd
These are the only 2 cases and there is only single box of balls.
Say, the balls you took out are: 1st ball is 2, 2nd ball is 4 and 3rd ball is 5
This case is clearly different from: 1st ball is 5, 2nd ball is 4 and 3rd ball is 2
Let us understand why:
Take a different question: The balls you have are numbered 1 to 4. You need to pick 2 balls with replacement and get an odd sum. What is the probability?
Let us determine the total cases: We often use the logic that the number of ways of picking the first ball is 4 and that for the 2nd ball is 4
=> So total = 4 x 4 = 16 cases
Observe that as soon as we are doing 4 x 4, we are considering the order. Let's count:
11 12 13 14 21 22 23 24 31 32 33 34 41 42 43 44 => 16 cases
Thus, its natural, when we count the favorable cases, we must consider the order too:
Cases for odd: 12, 14, 21, 23, 32, 34, 41, 43 => 8 cases => p(odd) = 8/16 = 1/2
Cases for even: 11, 13, 22, 24, 31, 33, 42, 44 => 8 cases => p(even) = 8/16 = 1/2
When you pick 2 balls, the sum can either be even or odd. Thus, p(even sum) + p(odd sum) = 1 (satisfied)
If you wish to not consider the order, then the denominator will not be 16 either - finding the total cases as a selection is trickier than finding the cases as an ordered selection (arrangement)
Thus, coming back to our question too, we should consider the order. You have 50 odd and 50 even balls.
Favorable cases:
Case 1: All 3 numbers are odd: 50 * 50 * 50 = 125 * 1000
Or
Case 2: 2 are even and 1 odd (EEO or EOE or OEE => 3 cases): favorable cases = 3 * (50 * 50 * 50) = 3 * 125 * 1000 = 375 * 1000
=> Favorable cases = 125 * 1000 + 375 * 1000 = 1000 * 500
Total cases = 100 * 100 * 100 = 1000 * 1000
=> Probability = (1000*500) / (1000*1000) = 1/2
Alternative approach:Case 1: All 3 numbers are odd: probability = 50/100 * 50/100 * 50/100 = 1/8
Or
Case 2: 2 are even and 1 odd (EEO or EOE or OEE => 3 cases): probability = 3 * (50/100 * 50/00 * 50/100) = 3/8
=> Probability = 1/8 + 3/8 = 4/8 = 1/2
Option C _________________
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