Conditional Probabilty of picking no black ball with red mark

Question

CLICK HERE TO SEE THE QUESTION Conditional probabilty question please help me with this question

Answer 1

Basic approach. There are three possible cases that satisfy the antecedent—that at least one black ball is picked: BW, WB, and BB. (It is important to consider BW and WB distinct, or else you will undercount the cases.) Since the selections are done with replacement, the probability of BW is

$$ P(\text{BW}) = P(\text{B}) \times P(\text{W}) = \frac14 \times \frac34 = \frac{3}{16} $$

The joint probability that the sequence of selections is BW, and there is no red dot in the two selections, is

$$ P(\text{BW, no red dot}) = P(\text{B, no red dot}) \times P(\text{W, no red dot}) = \frac14 \times \frac45 \times \frac34 = \frac{3}{20} $$

Repeat this process for the sequences WB and BB, keeping in mind that for the sequence BB, both draws of a black ball have to avoid a red dot. Then your desired probability is

\begin{align} P(\text{no red dot} & \mid \text{at least one black ball}) \\ & = \frac{P(\text{no red dot and (BW or WB or BB)}}{P(\text{BW or WB or BB})} \\ & = \frac{P(\text{BW, no red dot}) + P(\text{WB, no red dot}) + P(\text{BB, no red dot})}{P(\text{BW}) + P(\text{WB}) + P(\text{BB})} \end{align}

Answer 2

The probability (of picking at least one black ball) is 1-probability (that no black balls are picked).

That's $1 - \displaystyle \left(\frac{3}{4}\right)^2=\frac{7}{16}$.

Call this P(Black).

The probability (of picking no black balls with a red dot AND that at least one black ball is picked) is the same as the probability of (picking no black balls with a red dot minus the probability that two white balls are chosen).

And that is $\displaystyle \left(\frac{95}{100}\right)^2-\frac{9}{16}=\frac{361}{400}-\frac{225}{400}=\frac{136}{400}=\frac{17}{50}$.

Call this P(B&R).

The conditional probability of P(R|B), that is R given that B happened, is $\displaystyle \frac{P(B\&R)}{P(B)}=\frac{17}{50}*\frac{16}{7}=\boxed{0.78}$

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So in short.

P(Black) = At least one black ball chosen

P(Red) = No reds chosen

P(Red|Black) = P(Red&Black)/P(Black)