Question
Suppose that $E$ and $F$ are mutually exclusive events of an experiment. Show that if independent trials of this experiment are performed, then $E$ will occur before $F$ with probability $P(E) /[P(E)+P(F)]$.
Step 1
We have two mutually exclusive events, \(E\) and \(F\), which means they cannot occur simultaneously. We want to find the probability that \(E\) occurs before \(F\) in independent trials of the experiment. Show more…
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Let $E$ and $F$ be mutually exclusive events in the sample space of an experiment. Suppose that the experiment is repeated until either event $E$ or event $F$ occurs. What does the sample space of this new super experiment look like? Show that the probability that event $E$ occurs before event $F$ is $P(E) /[P(E)+P(F)]$. Hint: Argue that the probability that the original experiment is performed $n$ times and $E$ appears on the $n$ th time is $P(E) \times(1-p)^{n-1}, n=1,2, \ldots$, where $p=P(E)+$ $P(F)$. Add these probabilities to get the desired answer.
Let $E$ and $F$ be mutually exclusive events in the sample space of an experiment. Suppose that the experiment is repeated until either event $E$ or event $F$ occurs. What does the sample space of this new super experiment look like? Show that the probability that event $E$ occurs before event $F$ is $P(E) /$ $[P(E)+P(F)]$ Hint: Argue that the probability that the original experiment is performed $n$ times and $E$ appears on the $n$ th time is $P(E) \times(1-p)^{n-1}, n=1,2, \ldots$, where $p=P(E)+P(F)$. Add these probabilities to get the desired answer.
Let $E$ and $F$ be mutually exclusive events in the sample space of an experiment. Suppose that the experiment is repeated until either event $E$ or event $F$ occurs. What does the sample space of this new super experiment look like? Show that the probability that event $E$ occurs before event $F$ is $P(E) /[P(E)+P(F)]$ Hint: Argue that the probability that the original experiment is performed $n$ times and $E$ appears on the $n$ th time is $P(E) \times(1-p)^{n-1}$ $n=1,2, \ldots$, where $p=P(E)+P(F) .$ Add these probabilities to get the desired answer.
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