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Work Shown:
Define the following events
Draw out a probability tree. See diagram below.
From the tree diagram, we can figure out that P(L) = 0.44 which is the probability of being late. Note how I considered the two probabilities P(B & L) and P(S & L)
The events (B & L) and (S & L) are mutually exclusive, so we can add them up. Combining them will have us find the overall probability of being late. For more information, check out the law of total probability.
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Now use a conditional probability formula to find...
P(B|L) = probability the man takes bus given he is late
P(B|L) = P(B & L)/P(L)
P(B|L) = 0.2/0.44
P(B|L) = 0.45454545454546 approximately
P(B|L) = 0.4545 rounding to four decimal places
note: when I write "P(B|L)", I am using a vertical bar. I'm not using an uppercase letter 'i' or a lowercase letter L.
If the man is late for work on a particular day, the probability that he took the bus will be 0.455.
Given information:
A man takes either a bus or the subway to work with probabilities 0.4 and 0.6, respectively.
When he takes the bus, he is late 50% of the days and when he takes the subway, he is late 40% of the days.
Let's define the events as:
Now, being late and choosing between bus or subway are two independent events.
So, the probability of being late and taking the bus will be,
[tex]P(L\cap B)=0.5\times 0.4\\=0.2[/tex]
The probability of being late and taking the subway will be,
[tex]P(L\cap S)=0.4\times 0.6\\=0.24[/tex]
So, the probability of being late will be,
[tex]P(L)=P(L\cap B)+P(L\cap S)\\=0.2+0.24\\=0.44[/tex]
Now, we need to calculate the conditional probability that the person took the bus, given that he was late,
[tex]P(B|L)=\dfrac{P(L\cap B)}{P(L)}\\=\dfrac{0.2}{0.44}\\=0.455[/tex]
Therefore, if the man is late for work on a particular day, the probability that he took the bus will be 0.455.
For more details, refer to the link:
https://brainly.com/question/10739947
