# home assignment

• Nov 8th 2007, 05:11 AM
MS BATOOL
home assignment

a. In which situation, we use Permutation and Combination.

b. An MBA applies for a job in two firms X and Y. The probability of being selected in firm X is 0.7 and being rejected at Y is 0.5. The probability of at least one of his applications being rejected is 0.6. What is the probability that he will be selected by one of the firm?

c. Husband and wife appear in an interview for two vacancies in the same post. The probability of husband’s selection is and that of wife is. What is the probability that:

1. Both of them will selected
2. Only one of them is selected
3. None of them is selected.
• Nov 9th 2007, 04:49 AM
Soroban
Hello, MS BATOOL!

Quote:

b) An MBA applies for a job in two firms X and Y.
The probability of being selected in firm X is 0.7 and being rejected at Y is 0.5.
The probability of at least one of his applications being rejected is 0.6.
What is the probability that he will be selected by one of the firm?

We want: . $P (X \vee Y)$

We are given: . $\begin{array}{ccccccc}P(X) &=& 0.7 &\quad & P(X') & = & 0.3\\
P(Y) & = & 0.5 & \quad & P(Y') & = & 0.5 \end{array}\qquad P(X' \vee Y') \:=\:0.6$

$\text{Then: }\;\underbrace{P(X' \vee Y')}\;=\;\underbrace{P(X')} + \underbrace{P(Y')} - P(X' \wedge Y')$
Hence: . . $0.6 \qquad\,=\quad 0.3 \;\;\,+\;\; 0.5 - \;P(X' \wedge Y')\qquad\Rightarrow\qquad P(X' \wedge Y') \:=\:0.2$

By DeMorgan's Law: . $P(X \vee Y) \;=\;P(X' \wedge Y')' \;=\;(0.2)' \;=\;\boxed{0.8}$

Quote:

c) Husband and wife appear in an interview for two vacancies in the same post.
The probability of husband’s selection is ?? and that of wife is ??

Also, can we assume that the events are independent?

What is the probability that:

1. Both of them will selected
2. Only one of them is selected
3. Neither of them is selected.

Let: . $\begin{array}{ccc}P(H) \:=\:h & \quad & P(H') \:=\:1-h \\ P(W) \:=\:w & \quad & P(W') \:=\:1-w\end{array}$

$1)\;P(\text{Both selected}) \;=\;P(H)\cdot P(W) \;=\;h\cdot w$

$2)\;P(\text{One selected}) \;=\;P(H)\cdot P(W') + (H')\cdot P(W) \;=\;h(1-w) + (1-h)w$

$3)\;P(\text{Neither selected}) \;=\;P(H')\cdot P(W') \;=\;(1-h)(1-w)$

• Nov 9th 2007, 06:53 AM
MS BATOOL
thanks
Respected Sir,
i hope you will be fine.......yeah we can assume that the events are independent........i just want to thank you for your usefull reply.
• Nov 9th 2007, 07:50 AM
MS BATOOL
hello Sir !
i want to ask you ....can you help me in other threads about Statistics which are sent by me.....actually today is the last day to submit the assignments......i will be very thankful to you for this act of kindness.