Question

Let E and F be two events, if $P(E|F)=0.5$, $P(E|\overline{F})=0.6$ and $P(F)=0.6$ then $P(E)$ equals

• A. 0.56
• B. 0.44
• C. 0.54
• D. 0.46

Hint:

The Theorem of Total Probability is like a weighted average. You are essentially finding the "average" chance of E happening, weighted by how likely F and not-F are to occur.

Answer 1

The Correct Option is A

To find the total probability of event $E$, we use the Theorem of Total Probability, which partitions the sample space into event $F$ and its complement $\overline{F}$.

Step 1: \color{redIdentify the Known Probabilities
We are given:
$P(F) = 0.6$
$P(E|F) = 0.5$
$P(E|\overline{F}) = 0.6$

Step 2: \color{redCalculate the Probability of the Complement
The sum of the probability of an event and its complement is always 1.
$P(\overline{F}) = 1 - P(F)$
$P(\overline{F}) = 1 - 0.6 = 0.4$.

Step 3: \color{redApply the Theorem of Total Probability
The total probability $P(E)$ is the sum of probabilities of $E$ occurring given $F$, and $E$ occurring given $\overline{F}$:
$P(E) = P(E|F)P(F) + P(E|\overline{F})P(\overline{F})$
$P(E) = (0.5)(0.6) + (0.6)(0.4)$.

Step 4: \color{redPerform the Final Calculation
Calculate the individual terms:
$(0.5 \times 0.6) = 0.30$
$(0.6 \times 0.4) = 0.24$
Summing them up:
$P(E) = 0.30 + 0.24 = 0.54$.
let's re-verify the values.
$P(E) = 0.30 + 0.24 = 0.54$.
Check the options: (1) 0.56, (2) 0.44, (3) 0.54, (4) 0.46.
The result is 0.54, which corresponds to Option (3).