* Man City star Sergio Aguero was dumped by girlfriend two days after crashSergio Aguero has been dumped by his pop star lover.The Manchester Citystar, 29, who broke a rib in a car crash , is said to have drifted apart from 31-year-old Karina Tejeda.
Karina, known as The Little Princess in their Argentinian homeland, and Aguero had been dating for four years.
But she was reportedly angry at him for attending a party with several models on a brief return to Argentina.View On WordPress
* #relationships #relationshipgoals #dating #girlfriend #love #beautiful #men #women #saints #packers #atlanta (at Atlanta,...
* Sir, did you just see that I dumped him? Attention, Cloud 9 employees. We have a nonverbal confirmation I dumped Garrett.
Can you get your girlfriend back after being dumped?
In GTA: San Andreas, you are able to date several of the female NPCs in the game. However, if you have enough bad dates or if you neglect them, your relationship meter drops to 0% and you are ...
find the probability of selecting exactly two women and at least two women when a six-person committee is selected from $7$ men and $4$ women?
A committee of six members is formed from a group of $7$ men and $4$
women. What is the probability that the committee contains
a. exactly two women?
b. at least two women?
My attempt : given $P(A) = 7/11$ and $P(B) = 4/11$
option a) probability that the commitee contains
exactly two women
$$= \frac{P(AB)}{P(B)} = \frac{P(A)P(B)}{P(B)} = \frac{\frac{7}{11} \cdot \frac{4}{11}}{\frac{4}{11}} = \frac{7}{11}$$
option b)the probability that the committee contains at least two women = $$\frac{P(AB)}{P(A \cup B)} = \frac{\frac{7}{11} \cdot \frac{4}{11}}{1- \frac{7}{11} \cdot \frac{4}{11}} = \frac{28}{93}$$
If my answer is correct or not, I would be more thankful to those rectifying my mistakes......
Why does “I don't have a girlfriend yet” sound better than “I haven't had a girlfriend yet?” [closed]
I feel like the first one sounds better but I can't explain why
Scifi short story of a woman who meets her husband's parents, but the husband and his father turn out to be monsters
The woman meets her in-laws, and the mother-in-law seems crazy at first. Then the mother-in-law slowly shows the girl that the two guys, the husband and father-in-law, are monsters or aliens or ...
Find the ways in which no husband can sit next to another husband
A group of 6 people consisting 3 married couples are to be seated together in a straight row. How many different ways are there of seating the 6 people if no husband is to sit next to another husband?
Since the husbands can't sit next to each other, I decided to place the wife in between the husband and arrange them.
H W H W H W 3! * 3! = 36. Apparently, the answer is 144. Can anyone explain what went wrong?
Correct logic of permuting 5 men and 5 women to find probability of different highest women rank
The problem reads like this: Problem Five men and $5$ women are ranked according to their scores on an examination. Assume that no two scores are alike and all $10!$ possible rankings are equally likely. Let $X$ denote the highest ranking achieved by a woman. (For instance, $X = 1$ if the top-ranked person is female.) Find $P(X = i),i = 1, 2, 3, . . ., 8, 9, 10$. Solution given was: $P(X=1)= \frac{5}{10}= \frac{1}{2}$ because there are 5 women and total of 10 to choose from $P(X=2)=\frac{5}{10}\times \frac{5}{9}=\frac{5}{18}$ because for rank1 thereare 5 men and total of 10 to choose from, for rank 2 (we want awoman) we still have 5 women but only a total of 9 to choosefrom. $P(X=3)=\frac{5}{10}\times \frac{4}{9}\times \frac{5}{8}=\frac{5}{36}$ $P(X=4)=\frac{5}{10}\times \frac{4}{9}\times \frac{3}{8}\times \frac{5}{7}=\frac{10}{168}$ $P(X=5)=\frac{5}{10}\times \frac{4}{9}\times \frac{3}{8}\times \frac{2}{7}\times \frac{5}{6}=\frac{5}{252}$ $P(X=6)=\frac{5}{10}\times \frac{4}{9}\times \frac{3}{8}\times \frac{2}{7}\times \frac{1}{6}\times \frac{5}{5}=\frac{1}{252}$ My solution was $P(X=1)=\frac{5\times 9!}{10!}=\frac{1}{2}$ because there are five women to occupy 1st rank and then there remained 9 which can permute in $9!$ ways. There are total $10!$ ways to permute $10$ people
$P(X=2)=\frac{5\times \binom{5}{4}\times 8!}{10!}=\frac{5}{18}$ because there are five women to occupy 2st rank. The 1st rank will be of a man. So we have to select $4$ out of $5$ men which will be ranked after $2$nd rank. These four men and remaining 4 women can be permuted in $8!$ ways.
$P(X=3)=\frac{5\times \binom{5}{3}\times 7!}{10!}=\frac{5}{72}$
$P(X=4)=\frac{5\times \binom{5}{2}\times 6!}{10!}=\frac{5}{504}$
$P(X=5)=\frac{5\times \binom{5}{1}\times 5!}{10!}=\frac{5}{6048}$
$P(X=6)=\frac{5\times \binom{5}{0}\times 4!}{10!}=\frac{1}{30240}$ Doubts Where my logic went wrong?
When I compared the two approach, I realized that the books solution is permuting ranks higher than the highest ranked girl, while my solution is permuting ranks lower than the highest ranked girl. So I was guessing what makes book solution not permute lower ranks and my solution not permuting higher ranks. Shouldn't we permute on both sides of highest ranked girl?
How do you get your ex boyfriend back if he already had a girlfriend when he was dating you and lied and didn't tell you he had another girlfriend already and dumped you for her?
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She has said she has no interest in you, move on.
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Movie is called. it's complicated
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