On any given day, a certain machine has either no malfunctions or
exactly one malfunction. The probability of malfunction on any given
day is 0.40. Machine malfunctions on different days are mutually
independent. Calculate the probability that the machine has its third
malfunction on the fifth day, given that the machine has not had three
malfunctions in the first three days.
Try:
Let $X$ be number of days until third malfunction (success). We see $X$ is binomial with $r=3$ and $p=0.4$. therefore,
$$ P(X = 5 | X > 3) = \frac{P(X=5)}{P(X>3)} = \frac{ {4 \choose 2 } 0.4^2 \cdot 0.6^3 }{1 – P(X=2)-P(X=1) } = \frac{0.20736}{1-0-0} = \boxed{0.20736} $$
is this correct?
Best Answer
The beginning is correct, as per Bayes theorem we have
$$ \mathrm{P}(X = 5|X > 3) = \frac{\mathrm{P}(X = 5, X > 3)}{\mathrm{P}(X > 3)} = \frac{\mathrm{P}(X = 5)}{\mathrm{P}{(X > 3)}}, $$
since $X > 3$ always true when $X = 5$.
Now, if $X = 5$, then we have two malfunctions during first $4$ days, and the third on fifth day, so
$$ \mathrm{P}(X = 5) = \binom{4}{2}\cdot 0.4^2\cdot 0.6^2\cdot 0.4. $$
Also we note that $\mathrm{P}(X > 3) = 1 - \sum_{x = 0}^{3}\mathrm{P}(X = x)$, and the only non-zero element is $\mathrm{P}(X = 3) = 0.4^3$. Final calculations give that $\mathrm{P}(X = 5|X > 3) \approx 0.14769$.