Question
If $\frac{1+3 p}{3}, \frac{1-p}{4}$ and $\frac{1-2 p}{2}$ are the probabilities of three mutually exclusive events, then the set of all values of $p$ is .................
Step 1
The probabilities of these events are given as $\frac{1+3p}{3}$, $\frac{1-p}{4}$, and $\frac{1-2p}{2}$ respectively. Show more…
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If $\frac{1}{4}(1+4 p), \frac{1}{4}(1-p)$ and $\frac{1}{2}(1-2 p)$ are the probabilities of three mutually exclusive events, then $2 p$ is
For mutually exclusive events $R_{1}, R_{2},$ and $R_{3},$ we have $P\left(R_{1}\right)=0.15, \quad P\left(R_{2}\right)=0.55, \quad$ and $P\left(R_{3}\right)=0.30 .$ Also, $P\left(Q | R_{1}\right)=0.40, P\left(Q | R_{2}\right)=0.20,$ and $P\left(Q | R_{3}\right)=0.70 .$ Find The following. $P\left(R_{3} | Q\right)$
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Bayes’ Theorem
For mutually exclusive events $R_{1}, R_{2},$ and $R_{3},$ we have $P\left(R_{1}\right)=0.15, \quad P\left(R_{2}\right)=0.55, \quad$ and $P\left(R_{3}\right)=0.30 .$ Also, $P\left(Q | R_{1}\right)=0.40, P\left(Q | R_{2}\right)=0.20,$ and $P\left(Q | R_{3}\right)=0.70 .$ Find The following. $P\left(R_{1}^{\prime} | Q\right)$
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