can someone help solve this quesiton.
a basket contains 5 apples, of which 1 is spoiled and the rest are good. If Henry is to select 2 apples from the basket simultaneously and at random, what is the probability that the 2 apples will include the spoiled apple?
1/5
3/10
2/5
1/2
3/5
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- ksc311
- RonPurewal
- Students
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- Joined: Tue Aug 14, 2007 8:23 am
three possible solutions:
(1) symmetry argument - the best, but don't worry about it if it doesn't come immediately to mind:
you're picking 2 apples out of five. therefore, because of the total symmetry of the situation (no apple is inherently any more or less likely to be picked than any other), each apple should logically have a 2/5 chance of being picked. this includes the bad apple, so the answer is 2/5.
(2) combinations:
this is a combination problem.
because you have to say 'yes' to 2 apples and 'no' to 3 apples, the TOTAL number of combinations (the denominator) equivalent to asking for the number of anagrams of 'yynnn'. that is (5!)/(2!3!) = 10. the numerator is the total # of combinations including the bad apple, which is 4 (the bad apple can be chosen in conjunction with each of the 4 other apples). therefore, the probability is 4/10 = 2/5.
(3) organized counting --> you should start doing this right away if you don't think of another viable solution quickly!
make a list of all possibilities, in alphabetical order, if the apples are a, b, c, d, x (where x is the bad one)
ab
ac
ad
ax
bc
bd
bx
cd
cx
dx
there are 10 combinations, of which 4 (ax, bx, cx, dx) include the bad apple. the probability is therefore 4/10 = 2/5.
(1) symmetry argument - the best, but don't worry about it if it doesn't come immediately to mind:
you're picking 2 apples out of five. therefore, because of the total symmetry of the situation (no apple is inherently any more or less likely to be picked than any other), each apple should logically have a 2/5 chance of being picked. this includes the bad apple, so the answer is 2/5.
(2) combinations:
this is a combination problem.
because you have to say 'yes' to 2 apples and 'no' to 3 apples, the TOTAL number of combinations (the denominator) equivalent to asking for the number of anagrams of 'yynnn'. that is (5!)/(2!3!) = 10. the numerator is the total # of combinations including the bad apple, which is 4 (the bad apple can be chosen in conjunction with each of the 4 other apples). therefore, the probability is 4/10 = 2/5.
(3) organized counting --> you should start doing this right away if you don't think of another viable solution quickly!
make a list of all possibilities, in alphabetical order, if the apples are a, b, c, d, x (where x is the bad one)
ab
ac
ad
ax
bc
bd
bx
cd
cx
dx
there are 10 combinations, of which 4 (ax, bx, cx, dx) include the bad apple. the probability is therefore 4/10 = 2/5.
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