Probability of Balls

Probability Level 1

If from each of the three boxes containing $3$ white and $1$ black, $2$ white and $2$ black, $1$ white and $3$ black balls, one ball is drawn at random. Then the probability that $2$ white and $1$ black balls will be drawn is?

\dfrac{1}{4}

\dfrac{1}{32}

\dfrac{32}{13}

\dfrac{13}{32}

4 solutions

Jaiveer Shekhawat
Dec 29, 2014

Nice solution, @jaiveer shekhawat +1 upvote. I liked the way you solved it.

Anuj Shikarkhane - 6 years, 5 months ago

Thanx.. DuDe ... got an upvote after a long time

jaiveer shekhawat - 6 years, 5 months ago

Yes, I did the same way.

Manish Mayank - 6 years, 5 months ago

Tim Test
Oct 31, 2018

This question provides a symmetric condition, which means that even when the white balls change colour into black and black change to white, the condition does not change at all.

Hence, the probability of taking 2 white balls and 1 black ball is same as taking 1 white balls and 2 black balls. Also, the probability of taking 3 white balls is same as taking 3 black balls.

$P(3W)+P(2W1B)+P(1W2B)+P(3B)=1$ $2P(3W)+2P(2W1B)=1$ $P(2W1B)=1/2 - P(3W)$ $=1/2 - (1/4)(2/4)(3/4)$ $=13/32$

Very elegant, much better than actually listing all the ways.

Zyberg NEE - 2 years ago

Brock Brown
Dec 31, 2014

from fractions import Fraction as frac
level = {0:['w','w','w','b']}
cur_lvl = 1
while cur_lvl <= 2:
    level[cur_lvl] = []
    for s in level[cur_lvl-1]:
        if cur_lvl == 1:
            for c in 'wwbb':
                level[cur_lvl].append(s+c)
        elif cur_lvl == 2:
            for c in 'wbbb':
                level[cur_lvl].append(s+c)
    cur_lvl += 1
count = 0
for balls in level[2]:
    if balls.count('w') == 2:
        if balls.count('b') == 1:
            count += 1
print "Answer:", frac(count,len(level[2]))

Lam Vo Nhat
Apr 16, 2020

Probability of Balls

4 solutions

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