Bob can decide to go to work either by car, bus, or commuter train. if he decides to go by car, there is a 50% chance he will be late due to traffic. if he goes by bus, which has special reserved lanes but is sometimes overcrowded, the probability of being late is only 20%. the commuter train, there is a 1% chance that he is late, but is more expensive than the bus. what is the probability that bob drove to work today given he arrives late to work?

The probability is 0.7043

Step-by-step explanation:

Let's call:

C: Bob goes to work by car

B: Bob goes to work by bus

T: Bob goes to work by Train

L: Bob arrives late to work

The probability that Bob drove to work today given that he arrives late to work is:

P(C/L) = P(C∩L)/P(L)

Where P(L) = P(C∩L) + P(B∩L) + P(T∩L)

So, the probability P(C∩L) that Bob goes to work by car and arrives late to work is calculate as:

P(C∩L)=(1/3)*(0.5) = (1/6) = 0.1667

Because 1/3 is the probability that Bob choose the car and 0.5 is the probability that he will be late at work given that he choose a car.

At the same way, the probability P(B∩L) that Bob goes to work by bus and arrives late to work  and the probability P(T∩L) that Bob goes to work by Train and arrives late to work are calculate as:

P(B∩L)=(1/3)*(0.2) = (1/15) = 0.0667

P(T∩L)=(1/3)*(0.01) = (1/300) = 0.0033

Then P(L) is calculated as:

P(L) = 0.1667 + 0.0666 + 0.0033

P(L) = 0.2367

Finally, The probability P(C/L) that Bob drove to work today given that he arrives late to work is:

P(C/L) = 0.1667/0.2367 = 0.7043

step-by-step explanation:

the original amount doesn't ''re just looking for the time it takes for 1/2 of the substance to we have

(1/2) =)     take the ln   of both sides

ln (1/2)   = )   and we can write

ln(1/2) = ) lne     and lne   = 1     so we have

ln(1/2) = -.1092t     divide both sides by -.1092

t = ln(1/2)/-.1092 = about 6.347 days = about 6.3 days (rounded)

[2, 2.5)

(2.5, 3]

step-by-step explanation:

i think its these two not sure [2, 2.5)

(2.5, 3]