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You randomly draw a marble from a bag of marbles that contains 77 blue marbles, 22 green marbles, and 11 red marble.
What is \text{P(not draw a blue marble})P(not draw a blue marble)?

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kudzordzifrancis
ANSWER

[tex]P( \bar B)= \frac{3}{10} [/tex]

EXPLANATION

The number of blue marbles is

n(B)=77

The number of Green marbles is

n(G)=22

The number of red marbles is

n(R)=11

The total number of marbles is

n(S)=110


The probability of picking a blue marble is

[tex]P(B)= \frac{n(B)}{n(S)} [/tex]



[tex]P(B)= \frac{77}{110} [/tex]


[tex]P(B)= \frac{7}{10} [/tex]


The probability of not selecting a blue marble is


[tex]P( \bar B)= 1 - P(B).[/tex]



[tex]P( \bar B)= 1 - \frac{7}{10} [/tex]


[tex]P( \bar B)= \frac{3}{10} [/tex]
JeanaShupp

Answer: [tex]\dfrac{3}{10}[/tex]

Step-by-step explanation:

Given: The number of blue marbles = 77

The total number of blue marbles = [tex]77+22+11=110[/tex]

Now, the probability of choosing a blue marble is given by :-

[tex]\text{P( draw a blue)}=\dfrac{\text{Number of blue marbles}}{\text{ Total number of marbles}}\\\\\Rightarrow\ \text{P(draw a blue)}=\dfrac{77}{110}=\dfrac{7}{10}[/tex]

Now, the probability of not choosing a blue marble is given by :-

[tex]\text{P(not draw a blue)}=1-\text{P( draw a blue)}\\\\\Rightarrow\ \text{P(not draw a blue)}= 1-\dfrac{7}{10}\\\\\Rightarrow\ \text{P(not draw a blue)}=\dfrac{3}{10}[/tex]

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