"1) P(at least one M&M is brown)
= 1 - P(no M&M is brown)
= 1 - (0.87) = 0.13"
is correct for picking one brown.
For picking at least one brown, you can proceed as in the driver's problem, at least one driver ...
In this case, if you pick two, and both of them are NOT brown, the probability is
P(not brown)*P(not brown)
=0.87²
=0.7569
Therefore the probability of choosing AT LEAST one brown is
1 - P(not brown)*P(not brown)
= 0.2431
The probability of choosing all (both) blue is
P(blue)*P(blue)
=0.24*0.24
=0.0576
Therefore the probability of NOT choosing ALL blues
=1-P(blue)*P(blue)
=1-0.0575
=0.9425
You pick 2 M&Ms from a large bag. What is the probability that 1) at least one is brown? 2) not all are blue?
brown=13%
yellow=14%
red=13%
blue=24%
orange=20%
green=16%
1) P(at least one M&M is brown)
= 1 - P(no M&M is brown)
= 1 - (0.87) = 0.13
Is this correct?
2) P(not all M&Ms are blue)
= 1 - P(all M&Ms are blue)???
1 answer
1 answer