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I'm a sample bag of m&m's there are 5 brown, 6 yellow, 4 blue, 3 green, and 2 orange. What is the probability of getting 3 yellow m&m's if 3 are taken at a time?

Respuesta :

There are a total of 5+6+4+3+2=20 M&Ms in the bag.  Since there are 6 yellow M&Ms in the bag, there's a 6/20 probability that the first M&M will be yellow.  Then, there are 5 more yellow M&Ms and 19 more total M&Ms in the bag, so there is a 5/19 chance the next M&M will be yellow.  Finally, there are 4 more yellow M&Ms in the bag and 18 total M&Ms, so there is a 4/18 chance the last M&M will be yellow.

Multiplying these together, we have:

(6*5*4)/(20*19*18)

This simplifies to:

1/57

Therefore, there is a 1/57 probability that all M&Ms will be yellow.
AL2006
There are (5+6+4+3+2)=20 M&Ms in the bag.
The probability of the first one being yellow is 6 out of 20 = 6/20 .
If it is then the probability of the second one being yellow is 5 out of 19 .
If it is then the probability of the third one being yellow is 4 out of 18 .

The probability of all three being yellow is then

       (6/20) x (5/19) x (4/18) = 120/6,840  =  3/171 = 1/57  =  1.75% (rounded)

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