© All Rights reserved @ LearnWithDash
Step 1: Define the Events
Let:
$X$ be the result of a fair die roll, which can be 1, 2, 3, 4, 5, or 6.
$W$ be the event that all the balls drawn are white.
We want to find $P(W)$, the probability that all the drawn balls are white.
Step 2: Use the Law of Total Probability
Since $X$ takes values 1 through 6 with equal probability,
$P(X = x) = \frac{1}{6}$ for $x \in \{1,2,3,4,5,6\}.$
By the law of total probability:
$$
P(W) = \sum_{x=1}^{6} P(W \mid X = x)\,P(X = x) = \frac{1}{6} \sum_{x=1}^{6} P(W \mid X = x).
$$
Step 3: Find $P(W \mid X = x)$ for Each $x$
The bag contains 6 white and 4 black balls (10 in total). The probability of drawing $x$ white balls out of $x$ draws (with no replacement) is:
$$
P(W \mid X = x) = \frac{\binom{6}{x}}{\binom{10}{x}},
$$
valid for $x \le 6$ (since there are only 6 white balls). We compute this for $x = 1$ to 6:
$x = 1: \; P(W \mid X = 1) = \frac{\binom{6}{1}}{\binom{10}{1}} = \frac{6}{10} = \frac{3}{5}.$
$x = 2: \; P(W \mid X = 2) = \frac{\binom{6}{2}}{\binom{10}{2}} = \frac{15}{45} = \frac{1}{3}.$
$x = 3: \; P(W \mid X = 3) = \frac{\binom{6}{3}}{\binom{10}{3}} = \frac{20}{120} = \frac{1}{6}.$
$x = 4: \; P(W \mid X = 4) = \frac{\binom{6}{4}}{\binom{10}{4}} = \frac{15}{210} = \frac{1}{14}.$
$x = 5: \; P(W \mid X = 5) = \frac{\binom{6}{5}}{\binom{10}{5}} = \frac{6}{252} = \frac{1}{42}.$
$x = 6: \; P(W \mid X = 6) = \frac{\binom{6}{6}}{\binom{10}{6}} = \frac{1}{210}.$
Step 4: Sum Over All Possible $x$
Summing these conditional probabilities:
$$
\sum_{x=1}^{6} P(W \mid X = x)
= \frac{3}{5} + \frac{1}{3} + \frac{1}{6} + \frac{1}{14} + \frac{1}{42} + \frac{1}{210}.
$$
Converting each term to a common denominator of 210:
$\frac{3}{5} = \frac{126}{210}$
$\frac{1}{3} = \frac{70}{210}$
$\frac{1}{6} = \frac{35}{210}$
$\frac{1}{14} = \frac{15}{210}$
$\frac{1}{42} = \frac{5}{210}$
$\frac{1}{210} = \frac{1}{210}$
Summing the numerators:
$$
126 + 70 + 35 + 15 + 5 + 1 = 252.
$$
Hence,
$$
\sum_{x=1}^{6} P(W \mid X = x) = \frac{252}{210} = \frac{42}{35}.
$$
Therefore,
$$
P(W) = \frac{1}{6} \times \frac{42}{35} = \frac{42}{210} = \frac{1}{5}.
$$
Step 5: Final Answer
The probability that all the drawn balls are white is
$$
\boxed{\frac{1}{5}}.
$$
