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Bunuel
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Bunuel
9 balls labelled 1 - 9 are placed within a jar and two are pulled at random without replacement. What is the probability that the sum of the numbers on those two balls will be even?

A. 16/81
B. 25/81
C. 4/9
D. 41/81
E. 5/9


Total ways to select 2 balls from the jar = 9p2 = 9!/7! = 9 x 8 = 72

Favorable cases = 5 x 4 + 4 x 3 = 32

P(E) = Facorable Cases / Total ways = 32 / 72 = 4/9. Answer is C.
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Bunuel
9 balls labelled 1 - 9 are placed within a jar and two are pulled at random without replacement. What is the probability that the sum of the numbers on those two balls will be even?

A. 16/81
B. 25/81
C. 4/9
D. 41/81
E. 5/9

I cant understand why am i wrong.

favourable outcomes

4 evens and 5 odds
hence selected 4E + 4E = 8
and 5O+ 5O = 10

hence favourable outcomes = 18

Now from 9 balls, we have to select 2 without replacement hence 9C2= 36

So why isn't the answer 1/2
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Bunuel
9 balls labelled 1 - 9 are placed within a jar and two are pulled at random without replacement. What is the probability that the sum of the numbers on those two balls will be even?

A. 16/81
B. 25/81
C. 4/9
D. 41/81
E. 5/9

I cant understand why am i wrong.

favourable outcomes

4 evens and 5 odds
hence selected 4E + 4E = 8
and 5O+ 5O = 10

hence favourable outcomes = 18

Now from 9 balls, we have to select 2 without replacement hence 9C2= 36

So why isn't the answer 1/2

The issue is with how you calculated the number of ways to pick 2 odd and 2 even balls. The correct calculation is:

  • The number of ways to pick 2 odd-numbered balls from 5 is 5C2 = 10.
  • The number of ways to pick 2 even-numbered balls from 4 is 4C2 = 6.

Thus, the total favorable outcomes are 10 + 6 = 16.

The total ways to select 2 balls from 9 is 9C2 = 36.

The probability is therefore 16/36 = 4/9.
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