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Answer :
We know that the bag contains:
• 6 white counters,
,• 7 black counters,
,• 4 green counters,
,• 17 counters in total.
We define the events:
• W = draw a white counter,
,• B = draw a black counter,
,• G = draw a green counter.
,•
We have the following probabilities:
• P(W) = # white counters / total # of counters = 6/17,
,• P(B) = # black counters / total # of counters = 7/17,
,• P(G) = # green counters / total # of counters = 4/17,
(a) First, we compute:
P(W and G) = # white and green counters / total # of counters = 0/17 = 0.
The probability of drawing a white counter or a green counter is given by:
[tex]P(\text{W or }G)=P(W)+P(G)-P(W\text{ and G)}=\frac{6}{17}+\frac{4}{17}=\frac{10}{17}[/tex](b) First, we compute:
P(B and G) = # black and green counters / total # of counters = 0/17 = 0.
The probability of drawing a black counter or a green counter is given by::
[tex]P(B\text{ or }G)=P(B)+P(G)-P(B\text{ and G)}=\frac{7}{17}+\frac{4}{17}=\frac{11}{17}[/tex](c) The probability of not drawing a green counter is:
[tex]P(\text{not G)}=1-P(G)=1-\frac{4}{17}=\frac{17-4}{17}=\frac{13}{17}[/tex]Answers
• (a) P(W or G) = 10/17
,• (b) P(B or G) = 11/17
,• (c) P(not G) = 13/17
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