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March 6, 2015

March 6, 2015

Posted by **CM** on Wednesday, October 14, 2009 at 3:44pm.

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)???

- stats -
**MathMate**, Wednesday, October 14, 2009 at 4:49pm"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

**Answer this Question**

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