The probability that Jacqueline will be elected to the students’ council is 0.6, and the probability that she will be selected to represent her school in a public-speaking contest is 0.75. What is the probability that Jacqueline is either elected to the students’ council or picked for the public-speaking contest?

Question

anonymous · Answer

was thinking: 0.6*0.25 + 0.75*0.4 = 0.45
but the answer is 0.85

anonymous · Answer

assuming they are independent (big assumption) you take 
$$.6+.75-.6\times .75$$

anonymous · Answer

which is not .75

anonymous · Answer

i mean it is not .85

anonymous · Answer

hmm

anonymous · Answer

it is in fact .9

anonymous · Answer

i do know they are independant tho

anonymous · Answer

well fine.  then answer is .9

anonymous · Answer

if they are independent then 
$$P(A\cap B)=P(A)P(B)$$

anonymous · Answer

P( A intersection B) was given at 0.5 the previous question

anonymous · Answer

??

anonymous · Answer

in that case they are not independent

anonymous · Answer

The probability that Jacqueline will be elected to the students’ council is 0.6, and the probability that she will be selected to represent her school in a public-speaking contest is 0.75. The probability of Jacqueline achieving both of these goals is 0.5.

Sorry i didnt type some of the question

anonymous · Answer

because if they were independent then the probability of the intersection is .6*.75=.45

anonymous · Answer

ooooooooooooooooooh

anonymous · Answer

yes so they are independant

anonymous · Answer

no no no no sorry no

anonymous · Answer

IF they were independent then the probability of the intersection would be .45 but it is .5

anonymous · Answer

what you are supposed to do in this case is say .6+.75-.5=.85

anonymous · Answer

oh so they are dependant?

anonymous · Answer

oh yes

anonymous · Answer

omg wow

anonymous · Answer

easy to check. if they were independent then the probability of the intersection would be the product of the probabilities.  but is is not

anonymous · Answer

.6*.75=.45 not .5

anonymous · Answer

yes so they are not mutually exclusive

anonymous · Answer

be that as it may all you have to do is say 
$$P(A)+P(B)-P(A\cap B)=.6+.75-.5=.85$$

anonymous · Answer

well those are separate things.

anonymous · Answer

if they were mutually exclusive then 
$$P(A\cap B)=0$$

anonymous · Answer

so no they are not mutually exclusive, nor are they independent.  they are dependent

anonymous · Answer

ahh yes thx makes soo much more sense now

anonymous · Answer

yw