Probability help

[Fabio Newton]

3. The probability that a final year student will pass the certified public accountant (CPA) examination is 0.60. The probability that a final year student will pass the CPA examination and get a job offer is 0.40. Suppose the student just found out that they passed the CPA examination, what is the probability of them getting a job offer?


4. In a certain Pick 3 lottery game, players pick 3 numbers from 0 to 9. The cost of the game is $200. If the 3-digit number the player chooses (they must be in the same order) is drawn, then he/she wins $100,000. What is the expected value of this game?

Help please !!

 

[Deleted member 4993]

3. The probability that a final year student will pass the certified public accountant (CPA) examination is 0.60. The probability that a final year student will pass the CPA examination and get a job offer is 0.40. Suppose the student just found out that they passed the CPA examination, what is the probability of them getting a job offer?


4. In a certain Pick 3 lottery game, players pick 3 numbers from 0 to 9. The cost of the game is $200. If the 3-digit number the player chooses (they must be in the same order) is drawn, then he/she wins $100,000. What is the expected value of this game?

Help please !!

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[Fabio Newton]

3>>>>> P(A) = 0.60 and P( A n B) = 0.40
P(B) = equal 0.40÷ 0.60 = 0.666
Is this the correct answer and formula?

 

[pka]

3>>>>> P(A) = 0.60 and P( A n B) = 0.40
P(B) = equal 0.40÷ 0.60 = 0.666
Is this the correct answer and formula?

If we are using \(A\) for the event of passing and \(B\) for the event of getting a job offer then we use:
\(\mathcal{P}(B|A)=\dfrac{\mathcal{P}(B\cap A)}{\mathcal{P}(A)}\).

 

[Fabio Newton]

If we are using \(A\) for the event of passing and \(B\) for the event of getting a job offer then we use:
\(\mathcal{P}(B|A)=\dfrac{\mathcal{P}(B\cap A)}{\mathcal{P}(A)}\).

So it is indeed correct
P (B) is unkown and P (B n A)= 0.40
P(A) = 0.60
0.40÷0.60

 

[Fabio Newton]

Question 4>>>>>>>>
000-999 equal 1000 possible combination and the value of each game is $200
Therefore the value of the game is 1000×200 = 200,000
Please verify