1. A box contains 5 black, 7 red and 6 green balls. Three balls are drawn from this box one after the other without replacement. What is the probability that the three balls are
a. all black balls
b. of different colors
c. two black and one green black.
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Answer:
Hypergeometrical distribution!
In the denominator you just put the fact that you are drawing two balls from 5+3+7=15 balls: 15 over 2.
In the numerator you put three terms, referring to either the two balls being green (2,0,0), black (0,2,0) or red (0,0,2). So 5 over 2 (times 3 over 0 times 7 over zero but they are both one) plus 3 over 2 plus 7 over 2.
p=(52)(30)(70)+(50)(32)(70)+(50)(30)(72)(152)=(52)+(32)+(72)(152)=10+3+21105≈32,4%
Step-by-step explanation:
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