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Suppose a bag has 6 red balls and 9 blue balls what is the probability od choosing 2 red balls successfully at random without replacement



Answer :

Answer:

[tex]\text{the probability of choosing 2 red balls without replacement}=\displaystyle\bf\frac{1}{7}[/tex]

Step-by-step explanation:

We can find the probability of choosing 2 red balls without replacement by using the probability formula:

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

where:

  • [tex]P(A) = \text{probability of event A}[/tex]
  • [tex]n(A) = \text{total outcomes of event A}[/tex]
  • [tex]n(S) = \text{total outcomes of all possibilities}[/tex]

Let:

  • [tex]A=\text{choosing a red ball on the 1st turn}[/tex]
  • [tex]B=\text{drawing a red ball on the 2nd turn}[/tex]

1st turn:

  • [tex]n(S)=6+9=15\ \text{(total number of balls)}[/tex]
  • [tex]n(A)=6\ \text{(number of red balls)}[/tex]

Then:

[tex]\begin{aligned}P(A)&=\frac{n(A)}{n(S)} \\\\&=\frac{6}{15} \\\\&=\frac{2}{5} \end{aligned}[/tex]

2nd turn:

  • [tex]n(S)=15-1=14\ \text{(1 ball is taken on the 1st turn)}[/tex]
  • [tex]n(B)=6-1=5\ \text{(1 red ball is taken on the 1st turn)}[/tex]

Then:

[tex]\begin{aligned}P(B)&=\frac{n(B)}{n(S)} \\\\&=\frac{5}{14} \end{aligned}[/tex]

Probability of choosing 2 red balls without replacement:

[tex]\begin{aligned}P(A\cap B)&=P(A)\times P(B)\\\\&=\frac{2}{5} \times\frac{5}{14} \\\\&=\bf\frac{1}{7} \end{aligned}[/tex]

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