P(A|B)<span>P(A intersect B) = 0.2 = P( B intersect A)
</span>A) P(A intersect B) = <span>P(A|B)*P(B)
Replacing the known vallues:
0.2=</span><span>P(A|B)*0.5
Solving for </span><span>P(A|B):
0.2/0.5=</span><span>P(A|B)*0.5/0.5
0.4=</span><span>P(A|B)
</span><span>P(A|B)=0.4
</span>
B) P(B intersect A) = P(B|A)*P(A)
Replacing the known vallues:
0.2=P(B|A)*0.6
Solving for P(B|A):
0.2/0.6=P(B|A)*0.6/0.6
2/6=P(B|A)
1/3=P(B|A)
P(B|A)=1/3
Answer:
252
Step-by-step explanation:
To be divisible by 3, it's digits have to add to a number that is a multiple of 3.
To be divisible by 4 its last 2 digits have to be divisible by 3.
So let's start with 1x1 which won't work because 1x1 is odd. so let's go to 2x2 and see what happens.
212 that's divisible by 4 but not 3
222 divisible by 3 but not 4
232 divisible by 4 but not 3
242 not divisible by either one.
252 I think this might be your answer
The digits add up to 9 which is a multiple of 3 and the last 2 digits are divisible by 4
Answer:
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The answer to the following question is letter D
