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There is a random variable $X$ that takes integer values from 1 to 20.

An event $A = \{10 \leq X \leq 19\}$ counts as a success.

An event $B = \{2 \leq X \leq 9\}$ counts as a failure.

An event $C = \{X = 20\}$ counts as three successes.

An event $D = \{X = 1\}$ counts as two failures.

What is the probability that three successes happen before three failures?

My solution

The probability that there are at least three failures equals $p=P(D)^2+P(D)P(B)+P(B)^3=\frac{1}{20}\frac{1}{20}+\frac{1}{20}\frac{8}{20}+\left(\frac{8}{20}\right)^3=\frac{173}{2000}$

I then count the probability of zero, one and two successes.

$P\{\text{there are no successes}\}=p=\frac{173}{2000}$

$P\{\text{there is only one success}\}=p*P\{A\}=\frac{173}{2000}\frac{1}{2}$

$P\{\text{there are only two successes}\}=p*P\{A\}^2=\frac{173}{2000}\frac{1}{4}$

So, my probability equals

$P(Q)=1 - \frac{173}{2000}\frac{7}{4}=\frac{6769}{8000}\approx84.86\%$

Am I right?

1 Answers 1

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Probabilities:

  • Single Success: $1/2$
  • Triple Success: $1/20$
  • Single Failure: $2/5$
  • Double Failure: $1/20$

To get three successes before three failures you must obtain one of the following disjoint sequences:

$$\boxed{\begin{array}{l:l} \text{A triple success.}&\tfrac 1{20}\\ \hdashline \text{ Three single successes }&{(\tfrac 12)}^3\\ \hdashline \text{A single failure and two successes, then a single success.} &\tbinom{3}{1}(\tfrac{2}{5}){(\tfrac{1}{2})}^3\\ \hdashline \text{ A single failure, then a triple success.}& \\ \hdashline \text{ } & \\ \hdashline \text{ } & \\ \hdashline \text{ }& \\ \hdashline \text{ }& \end{array}}$$

Complete the table, sum and simplify.