A bag contains some white and black balls . The probability of picking two white balls one after other without replacement from that bag is 14/33. Then what will be the probability of picking two black balls from that bag if bag can hold maximum 15 balls only?

Answers

Answer 1

Answer:

2/35

Step-by-step explanation:

Let w and b be the numbers of white and black balls in the bag respectively.

So, the total numbers of the balls in the bag is

[tex]n=w+b\;\cdots(i)[/tex]

As the bag can hold maximum 15 balls only, so

[tex]n\leq15 \;\cdots(ii)[/tex]

Probability of picking two white balls one after other without replacement

=Probability of the first ball to be white and the probability of second ball to be white

=(Probability of picking first white balls) x( Probability of picking 2nd white ball)

Here, the probability of picking the first white ball [tex]=\frac{w}{n}[/tex]

After picking the first ball, the remaining

white ball in the bag [tex]= w-1[/tex]

and the remaining total balls in the bag [tex]=n-1[/tex]

So, the probability of picking the second white ball [tex]=\frac{w-1}{n-1}[/tex]

Given that, the probability of picking two white balls one after other without replacement is  14/33.

[tex]\Rightarrow \frac{w}{n} \times \frac{w-1}{n-1}=\frac{14}{33}[/tex]

[tex]\Rightarrow \frac{w(w-1)}{n(n-1)} =\frac{14}{33}[/tex]

Here, [tex]w[/tex] and [tex]n[/tex] are counting numbers (integers) and 14 and 33 are co-primes.

Let, [tex]\alpha[/tex] be the common factor of the numbers [tex]w(w-1)[/tex] (numerator) and [tex]n(n-1)[/tex] (denominator), so

[tex]\frac{w(w-1)}{n(n-1)} =\frac{14\alpha}{33\alpha}[/tex]

[tex]\Rightarrow w(w-1)=14\alpha\cdots(iii)[/tex]

And [tex]n(n-1)=33\alpha.[/tex]

As from eq. (ii), [tex]n\leq 15[/tex], so, the possible value of [tex]\alpha[/tex] for which multiplication od two consecutive positive integers (n and n-1) is [tex]33\alpha[/tex] is 4.

[tex]n(n-1)=11\times (3\alpha)[/tex]

[tex]\Rightarrow n(n-1)=12\times11[/tex] [as [tex]\alpha=4[/tex]]

[tex]\Rightarrow n=12[/tex]

So, the number of total balls =12

From equation (iv)

[tex]w(w-1)=7\times (2\alpha)[/tex]

[tex]\Rightarrow w(w-1)=8\times7[/tex]

[tex]\Rightarrow w=8[/tex]

So, the number of white balls =8

From equations (i), the number of black balls =12-8=4

In the similar way, the required probability of picking two black balls one after other in the same way (i.e without replacement) is

[tex]=\frac{b}{n} \times \frac{b-1}{n-1}[/tex]

[tex]= \frac{4}{15} \times \frac{4-1}{15-1}[/tex]

[tex]= \frac{4}{15} \times \frac{3}{14}[/tex]

[tex]= \frac{2}{35}[/tex]

probability of picking two black balls one after other without replacement is 2/35.


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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Answer:

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Answers

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Answers

Answer:

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Answers

Answer:

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Data;

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The amount Sangeetha got

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Answers

Answer:

261

Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Answers

Answer:

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Answer:

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Answers

The scatterplot shows that the prediction of the height in centimeters of a student with a weight of 64 kilograms is 156cm.

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It should be noted that a scatterplot simply means a type of data display which shows the relationship between two numerical variables.

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The values of the radius and volume are placed in the formula to get the volume of the cylinder

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Answer:

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Step-by-step explanation:

So long as you have the volume and radius of a cylinder, it should be easy!

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