Probability of equal # of blue and red fish chosen when drawing without replacement
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The problem is such
One diagram of the possible situations is as follows:
M M F F , $P(M=2,F=2)$
Need to find , $ P(M=2,F=2)$, So:
-# of ways to pick four fish from the pond = $C_2^{10}C_2^{8}=45*28=1260$
-# of ways to pick 2 male and female fish from the pond = $(10*9)+(8*7)=90+56$
So the probability is $frac{146}{1260}=0.11589$, which seems a bit too small; shouldn't it be closer to 1 as what other cases would make up for the remaining 1114(=1260-146) cases. I am guessing that I went wrong with finding the possible ways as 146. And I am not sure if I need to consider if there was only 1 male and female fish and 0 male and female fish. Cause it doesn't make sense that when you draw 4 fish u get none or only 2 fish.
probability proof-verification
asked Nov 15 at 10:01
glockm15
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The problem is such
One diagram of the possible situations is as follows:
M M F F , $P(M=2,F=2)$
Need to find , $ P(M=2,F=2)$, So:
-# of ways to pick four fish from the pond = $C_2^{10}C_2^{8}=45*28=1260$
-# of ways to pick 2 male and female fish from the pond = $(10*9)+(8*7)=90+56$
So the probability is $frac{146}{1260}=0.11589$, which seems a bit too small; shouldn't it be closer to 1 as what other cases would make up for the remaining 1114(=1260-146) cases. I am guessing that I went wrong with finding the possible ways as 146. And I am not sure if I need to consider if there was only 1 male and female fish and 0 male and female fish. Cause it doesn't make sense that when you draw 4 fish u get none or only 2 fish.
probability proof-verification
asked Nov 15 at 10:01
glockm15
32519
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The problem is such
One diagram of the possible situations is as follows:
M M F F , $P(M=2,F=2)$
Need to find , $ P(M=2,F=2)$, So:
-# of ways to pick four fish from the pond = $C_2^{10}C_2^{8}=45*28=1260$
-# of ways to pick 2 male and female fish from the pond = $(10*9)+(8*7)=90+56$
So the probability is $frac{146}{1260}=0.11589$, which seems a bit too small; shouldn't it be closer to 1 as what other cases would make up for the remaining 1114(=1260-146) cases. I am guessing that I went wrong with finding the possible ways as 146. And I am not sure if I need to consider if there was only 1 male and female fish and 0 male and female fish. Cause it doesn't make sense that when you draw 4 fish u get none or only 2 fish.
probability proof-verification
asked Nov 15 at 10:01
glockm15
32519
The problem is such
One diagram of the possible situations is as follows:
M M F F , $P(M=2,F=2)$
Need to find , $ P(M=2,F=2)$, So:
-# of ways to pick four fish from the pond = $C_2^{10}C_2^{8}=45*28=1260$
-# of ways to pick 2 male and female fish from the pond = $(10*9)+(8*7)=90+56$
So the probability is $frac{146}{1260}=0.11589$, which seems a bit too small; shouldn't it be closer to 1 as what other cases would make up for the remaining 1114(=1260-146) cases. I am guessing that I went wrong with finding the possible ways as 146. And I am not sure if I need to consider if there was only 1 male and female fish and 0 male and female fish. Cause it doesn't make sense that when you draw 4 fish u get none or only 2 fish.
probability proof-verification
probability proof-verification
asked Nov 15 at 10:01
glockm15
32519
asked Nov 15 at 10:01
glockm15
32519
asked Nov 15 at 10:01
glockm15
32519
asked Nov 15 at 10:01
glockm15
32519
asked Nov 15 at 10:01
glockm15
32519
32519
add a comment |
add a comment |
1 Answer
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You have to consider whole items.
Now Total = 18C4 = 3060
Like you said, ways to pick 2 male and female fish from the pond = 10C2 * 8C2 = 1260
So the probability is 1260 ⁄ 3060=0.4117
answered Nov 15 at 11:13
ImNv
16
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
You have to consider whole items.
Now Total = 18C4 = 3060
Like you said, ways to pick 2 male and female fish from the pond = 10C2 * 8C2 = 1260
So the probability is 1260 ⁄ 3060=0.4117
answered Nov 15 at 11:13
ImNv
16
add a comment |
up vote
0
down vote
accepted
You have to consider whole items.
Now Total = 18C4 = 3060
Like you said, ways to pick 2 male and female fish from the pond = 10C2 * 8C2 = 1260
So the probability is 1260 ⁄ 3060=0.4117
answered Nov 15 at 11:13
ImNv
16
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
You have to consider whole items.
Now Total = 18C4 = 3060
Like you said, ways to pick 2 male and female fish from the pond = 10C2 * 8C2 = 1260
So the probability is 1260 ⁄ 3060=0.4117
answered Nov 15 at 11:13
ImNv
16
You have to consider whole items.
Now Total = 18C4 = 3060
Like you said, ways to pick 2 male and female fish from the pond = 10C2 * 8C2 = 1260
So the probability is 1260 ⁄ 3060=0.4117
answered Nov 15 at 11:13
ImNv
16
answered Nov 15 at 11:13
ImNv
16
answered Nov 15 at 11:13
ImNv
16
answered Nov 15 at 11:13
ImNv
16
16
add a comment |
add a comment |
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