probability help!

[martin1106]

 I don't understand how they derive this answer. Pls help me, thanks. 

[b^3]

The ways we can have 2 drivers absent in the next three days are:
1 driver absent on day 1 & 2     
1 driver absent on day 1 & 3     
1 driver absent on day 2 & 3     

2 drivers absent on day 1         
2 drivers absent on day 2         
2 drivers absent on day 3         

If we let Y be the number of drivers absent in the three days then

[fred42]

The distribution gives the probabilities of absent drivers on any one day, so, for 2 drivers to be absent on 3 days, we need:

Prob( exactly 2 drivers absent) = Prob(2 drivers absent on one day and none on the other 3) X3 +                                            Prob(1 driver absent on 2 days plus nine absent on the other day)X3
         

[martin1106]

thanks guys! XD

[BubbleWrapMan]

Egh, got ninja'd but I'll post it anyway:


The only ways you could have two drivers absent over 3 days is if:

Two absences occur on one day and none occur on the other two days (this can occur in = 3 ways: 002, 020, 200). This will have a probability of .

Or, one absence occurs on two days, and no absence occurs on the other day (this can occur in = 3 ways: 011, 101, 110). So this probability is .

Those two cases are mutually exclusive so you add their probabilities together to get the union.

[martin1106]

many thanks!