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mathmath333
 one year ago
Probablity question
mathmath333
 one year ago
Probablity question

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2\(\large \color{black}{\begin{align} & \normalsize \text{A man and woman appear in an interview for 2 vacancies}\hspace{.33em}\\~\\ & \normalsize \text{of same post.The probablity of man's selection is }\ \dfrac{1}{4} \text{and that of } \hspace{.33em}\\~\\ & \normalsize \text{woman's selection is }\ \dfrac{1}{3}\ \text{What is the probablity that none } \hspace{.33em}\\~\\ & \normalsize \text{of them will be selected} \hspace{.33em}\\~\\ & a.)\ \dfrac{1}{2} \hspace{.33em}\\~\\ & b.)\ \dfrac{1}{3} \hspace{.33em}\\~\\ & c.)\ \dfrac{2}{5} \hspace{.33em}\\~\\ & d.)\ \dfrac{3}{7} \hspace{.33em}\\~\\ \end{align}}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2same thing, use the fact that the events are independent : \[P(A'\cap B') = P(A')\times P(B')\]

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2is it also \([1P(\text{both will be selected)}]\)

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2hint: If 1/4 is P(selected), so \((1\frac{1}{4})\) is P(not selected). Similarly for the woman.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2nope, you need to subtract probabilities for exactly one of them is selected too

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2all that is a pain in the neck, so just use mathmate's hint

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2matemate what is ur full hint

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1and since they are independent events the probabilities are multiplied

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2My hint is just a followup to yours, lol. @ganeshie8

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2Use the formula given by @ganeshie8 My hint is to help you find P(A') and P(B'). A' is the complement of A. For example, A=P(selected)=1/4, then A'=P(not selected)=11/4=3/4. So do this for both man and woman, and use @ganeshie8 's hint.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2is it 1/2 final answer

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2How did you arrive at 1/2? Please show us some details to make sure.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2\(\large \color{black}{\begin{align} P(A'\cap B') &= P(A')\times P(B')\hspace{.33em}\\~\\ &= (1P(A))\times (1P(B))\hspace{.33em}\\~\\ &= (11/4)\times (11/3)\hspace{.33em}\\~\\ &= (3/4)\times (2/3)\hspace{.33em}\\~\\ &= (1/2)\hspace{.33em}\\~\\ \end{align}}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2just for more practice, maybe try working the probability using ur initial method

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2is it also \([1P(\text{both will be selected)}]\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2as i said earlier, that is wrong, here is the correct one : \(\begin{align}P(\text{none selected})&=1P(\text{both will be selected)}\\&P(\text{man selected, woman not selected)}\\ & P(\text{man not selected, woman selected)}\end{align} \)
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