mathmath333
  • mathmath333
Probablity question
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{A man and woman appear in an interview for 2 vacancies}\hspace{.33em}\\~\\ & \normalsize \text{of same post.The probablity of man's selection is }\ \dfrac{1}{4} \text{and that of } \hspace{.33em}\\~\\ & \normalsize \text{woman's selection is }\ \dfrac{1}{3}\ \text{What is the probablity that none } \hspace{.33em}\\~\\ & \normalsize \text{of them will be selected} \hspace{.33em}\\~\\ & a.)\ \dfrac{1}{2} \hspace{.33em}\\~\\ & b.)\ \dfrac{1}{3} \hspace{.33em}\\~\\ & c.)\ \dfrac{2}{5} \hspace{.33em}\\~\\ & d.)\ \dfrac{3}{7} \hspace{.33em}\\~\\ \end{align}}\)
ganeshie8
  • ganeshie8
same thing, use the fact that the events are independent : \[P(A'\cap B') = P(A')\times P(B')\]
mathmath333
  • mathmath333
is it also \([1-P(\text{both will be selected)}]\)

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More answers

mathmate
  • mathmate
hint: If 1/4 is P(selected), so \((1-\frac{1}{4})\) is P(not selected). Similarly for the woman.
ganeshie8
  • ganeshie8
nope, you need to subtract probabilities for exactly one of them is selected too
ganeshie8
  • ganeshie8
all that is a pain in the neck, so just use mathmate's hint
mathmath333
  • mathmath333
matemate what is ur full hint
welshfella
  • welshfella
and since they are independent events the probabilities are multiplied
mathmate
  • mathmate
My hint is just a follow-up to yours, lol. @ganeshie8
mathmath333
  • mathmath333
oh so it's same
mathmath333
  • mathmath333
1/2
mathmate
  • mathmate
Use 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)=1-1/4=3/4. So do this for both man and woman, and use @ganeshie8 's hint.
mathmath333
  • mathmath333
is it 1/2 final answer
mathmate
  • mathmate
How did you arrive at 1/2? Please show us some details to make sure.
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} P(A'\cap B') &= P(A')\times P(B')\hspace{.33em}\\~\\ &= (1-P(A))\times (1-P(B))\hspace{.33em}\\~\\ &= (1-1/4)\times (1-1/3)\hspace{.33em}\\~\\ &= (3/4)\times (2/3)\hspace{.33em}\\~\\ &= (1/2)\hspace{.33em}\\~\\ \end{align}}\)
ganeshie8
  • ganeshie8
looks good to me!
mathmath333
  • mathmath333
thnks
ganeshie8
  • ganeshie8
just for more practice, maybe try working the probability using ur initial method
ganeshie8
  • ganeshie8
is it also \([1-P(\text{both will be selected)}]\)
ganeshie8
  • ganeshie8
as i said earlier, that is wrong, here is the correct one : \(\begin{align}P(\text{none selected})&=1-P(\text{both will be selected)}\\&-P(\text{man selected, woman not selected)}\\ &- P(\text{man not selected, woman selected)}\end{align} \)

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