Two balls are placed in a box according to the following process: 1. A fair coin is tossed. A white ball is placed in the box if a head occurs, otherwise a red ball is placed in the box. 2. The fair coin is tossed again. A red ball is placed in the box if a tail occurs, otherwise a white ball is placed in the box. Balls are drawn from the box three times in succession (always with replacing the drawn ball back in the box). On all three occasions a red ball is drawn from the box. There are two pieces of information available: the process by which balls are placed in the box and the 3 observations of a ball drawn from the box. Making use of all relevant available information, what is the probability that both balls in the box are red

Answer:

The probability that both balls in the box are red given that all three occasions a red ball is drawn from the box is [tex]\frac{4}{5}[/tex]

Step-by-step explanation:

Let RRR be the event in which red Balls are drawn from the box three times in succession

Let rr be the event that two red balls are placed in a box

Let rw be the event the first red ball was placed and then white ball was placed

Let wr be the event the first white ball was placed and then red ball was placed

Let ww be the event that two white balls are placed in a box

Now we know that :

[tex]P(rr)=P(rw)=P(wr)=P(ww)=\frac{1}{4}[/tex]

Probability of getting red Balls three times in succession given that both are placed red balls in box:

[tex]P(RRR|rr)=1[/tex]

Probability of getting red Balls three times in succession given that one ball is red and another is white in a box

[tex]P(RRR|wr)=P(RRR|rw)=\frac{1}{2^3}=\frac{1}{8}[/tex]

Probability of getting red Balls three times in succession given that both are placed white balls in box:

[tex]P(RRR|ww)=0[/tex]

Now we are supposed to find the probability that both balls in the box are red given that  all three occasions a red ball is drawn from the box i.e. P(rr|RRR)

So, we will use bayes theorem

[tex]P(rr|RRR)=\frac{P(RRR|rr) P(rr)}{P(RRR|rr) P(rr)+P(RRR|rw) P(rw)+P(RRR|wr) P(wr)+P(RRR|ww) P(ww)}\\P(rr|RRR)=\frac{1 \times \frac{1}{4}}{1 \times \frac{1}{4}+\frac{1}{8} \times \frac{1}{4}+\frac{1}{8} \times \frac{1}{4}+0 \times \frac{1}{4}}\\\\P(rr|RRR)=\frac{\frac{1}{4}}{\frac{5}{16}}\\\\P(rr|RRR)=\frac{4}{5}[/tex]

Hence the probability that both balls in the box are red given that all three occasions a red ball is drawn from the box is [tex]\frac{4}{5}[/tex]

Did I graph this correctly?

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