Two professors are applying for grants. Professor Jane has a probability of 0.61 of being funded. Professor Joe has probability 0.27 of being funded. Since the grants are submitted to two different federal agencies, assume the outcomes for each grant are independent. Give your answer to four decimal places. a. What is the probability that both professors get their grants funded? Give your answer to four decimal places

Accepted Solution

Answer:a. 0.1647;b. 0.7153;c. 0.4453.Step-by-step explanation:By definition, if two event [tex]A[/tex] and [tex]B[/tex] are independent, then[tex]P(A\cap B) = P(A) \cdot P(B)[/tex]. ([tex]P(A\cap B)[/tex] is the probability that the outcome of both event [tex]A[/tex] and event [tex]B[/tex] are true.) a.Since the outcome of these two events are independent, [tex]\begin{aligned}P(\texttt{Jane} \cap \texttt{Joe}) &= P(\texttt{Jane}) \cdot P(\texttt{Joe})\\ &= 0.61 \times 0.27 \\&= 0.1647\end{aligned}[/tex].b.The logic not operator [tex]\lnot[/tex] or the prime superscript [tex]^{\prime}[/tex] denotes that an event does not happen.  [tex]P(\texttt{Jane}^{\prime}) = 1 - P(\texttt{Jane}) = 1- 0.61 = 0.39[/tex].[tex]P(\texttt{Joe}^{\prime}) = 1 - P(\texttt{Joe}) = 1- 0.27 = 0.73[/tex].Since the two events [tex]\texttt{Jane}[/tex] and [tex]\texttt{Joe}[/tex], [tex]\texttt{Jane}^{\prime}[/tex] and [tex]\texttt{Joe}^{\prime}[/tex] are also independent. Probability that neither professor got funded:[tex]P(\texttt{Jane}^{\prime} \cap \texttt{Joe}^{\prime}) = P(\texttt{Jane}^{\prime}) \cdot P(\texttt{Joe}^{\prime}) = 0.39 \times 0.73 = 0.2847[/tex].Probability that at least one professor got funded- in other words, it is not true that neither professor got funded:[tex]P((\texttt{Jane}^{\prime} \cap \texttt{Joe}^{\prime})^{\prime}) = 1- P(\texttt{Jane}^{\prime} \cap \texttt{Joe}^{\prime}) = 0.7153[/tex].c.Similarly, since the two events [tex]\texttt{Jane}[/tex] and [tex]\texttt{Joe}[/tex], [tex]\texttt{Jane}[/tex] and [tex]\texttt{Joe}^{\prime}[/tex] are also independent. Probability that Jane but not Joe got funded:[tex]P(\texttt{Jane} \cap (\texttt{Joe}^{\prime})) = P(\texttt{Jane}) \cdot P(\texttt{Joe}^{\prime}) = 0.61 \times 0.73 = 0.4453[/tex].