Respuesta :
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.
