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) 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
1) To find the probability that at least one M&M is brown, we need to calculate the probability that no M&M is brown and subtract it from 1.
The probability of no M&M being brown can be calculated by multiplying the probabilities of both M&Ms being non-brown. Since we have two M&Ms and the probabilities are given as percentages, we need to convert the percentages to decimals before multiplying.
P(no M&M is brown) = 0.87 (probability of first M&M not being brown) * 0.87 (probability of second M&M not being brown)
= 0.87 * 0.87 = 0.7569
Therefore, the probability that at least one M&M is brown is:
P(at least one M&M is brown) = 1 - P(no M&M is brown)
= 1 - 0.7569
= 0.2431
So, your answer of 0.13 is incorrect. The correct probability is 0.2431.
2) To find the probability that not all M&Ms are blue, we can use the same approach as above. We need to calculate the probability of all M&Ms being blue and subtract it from 1.
The probability of all M&Ms being blue can be found by multiplying the probabilities of both M&Ms being blue. Again, we need to convert the percentages to decimals before multiplying.
P(all M&Ms are blue) = 0.24 (probability of first M&M being blue) * 0.24 (probability of second M&M being blue)
= 0.24 * 0.24 = 0.0576
Therefore, the probability that not all M&Ms are blue is:
P(not all M&Ms are blue) = 1 - P(all M&Ms are blue)
= 1 - 0.0576
= 0.9424
So, your suggestion of using 1 - P(all M&Ms are blue) to find the probability is correct. The correct probability is 0.9424.