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
As the bag can hold maximum 15 balls only, so
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
After picking the first ball, the remaining
white ball in the bag
and the remaining total balls in the bag
So, the probability of picking the second white ball
Given that, the probability of picking two white balls one after other without replacement is 14/33.
Here, and
are counting numbers (integers) and 14 and 33 are co-primes.
Let, be the common factor of the numbers
(numerator) and
(denominator), so
And
As from eq. (ii), , so, the possible value of
for which multiplication od two consecutive positive integers (n and n-1) is
is 4.
[as
]
So, the number of total balls =12
From equation (iv)
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
probability of picking two black balls one after other without replacement is 2/35.