Wu wins 30-hole match, reaches US Women's Amateur semis

Chia Yen Wu advanced to the semifinals of the U.S. Women's Amateur after rolling in a par putt on the 12th playoff hole to win the longest 18-hole match in … Click to Continue »
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  • Qatar Classic 2017 - Day SIX: QUARTER-FINALS Down to the last eight in Doha, with six of the top eight seeds, three former champions and four Egyptians still in contention for the 2017 title. [3] Mohamed Elshorbagy (Egy) 3-2 [5] Ali Farag (Egy)   11/7, 11/4, 9/11, 16/18, 11/8 (93m) What a match, what a match, what a match … and that makes six Qatar semis in a row for Shorbagy …   [1] Gregory Gaultier (Fra) 3-0 Diego Elias (Per)    11/2, 11/7, 11/5 (44m) A comfortable win for the world #1 and top seed in his comeback tournament, but another great week for Diego. Simon Rosner (Ger) 3-1 [6] Marwan Elshorbagy (Egy)     11/7, 9/11, 11/5, 11/3 (53m) The big German continues his impressive run, taking another Egyptian scalp to reach his first World Series semi-final …. [8] Tarek Momen (Egy) 3-1 [4] Nick Matthew (Eng)     11/5, 9/11, 11/7,  11/6 (43m) A second upset to finish the round as Momen reaches his first Qatar Classic semi-final.
  • Late substitute @MohamedDiame17 wins a corner & almost wins the match at the end of a creditable 1-1 draw with Liverpool that...
  • Golfers defy odds to hit back-to-back holes-in-one [ad_1] Media playback is unsupported on your device A pair of amateur golfers have defied astronomical odds by hitting consecutive holes-in-one. Jayne Mattey and Clair Shine achieved the feat on the 13th hole at the East Berkshire Golf Club in Crowthorne, Berkshire, on Saturday. The National Hole-In-One-Registry has calculated the odds of two players from the same foursome acing the same hole as… View On WordPress
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Wu wins 30-hole match, reaches US Women's Amateur semis
Chia Yen Wu advanced to the semifinals of the U.S. Women's Amateur after rolling in a par putt on the 12th playoff hole to win the longest 18-hole match in … Click to Continue »
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Bolt reaches end of hole but threads don't grip Is there anything I can use short of Helicoil, which I'm really hoping not to have to use, to make sure a bolt sits in its threaded socket in my engine block and torques to spec ? The problem that I ...
Using semicolons to create run-on sentences … what's the deal with semis? I often see people making sentences quite longer than I'm comfortable with, such as like this: The dog ran, the dog fell, the dog dwelled; the dog didn't wish to be a part of such a place in his ...
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......
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?
If staring at women is such a big aveira how come rabbis are permitted to give shiurim to women? There are many great rabbis that I know who also give classes to women. Every where else I see that guarding your eyes is of crucial importance, but how are rabbis who look directly at women for long ...
Guy who dates/hooks up with a lot of women, divided into parts by the women's names, was on Netflix around 2016-2017 It was on Netflix a while ago and I couldn't find it on there. Set in the US, there is this good looking but creepy player type guy who is always going out and is smoking and on his phone, he hooks up ...
Does an amateur's caddy get anything if the amateur wins the US Open? no
If an amateur wins a PGA tournament do they pay him? No, a term of accepting the sponsors invite is that he doesn't receive any official money, so no, he receives none of the tournament purse.
Can a amateur keep the money if he wins? No, when an amateur accepts an invitation to a tour event they forgo all official money.
Can an amateur declare being pro if he wins? No, his status must be declared prior to the start of the competition.
If an amateur wins the British open does he get the claret jug? Yes, an amateur would win the Claret Jug if he won the Open Champoinship. However, he would not win the prize money, as that would contavene his amateur status
How long is an amateur boxing match?
What happens to the money if an amateur wins the US Open golf tournament? The proffesional player who finishes 2nd wins the top prize.
What boxer knocked out his opponent in the first amateur boxing match?
If an amateur wins a PGA tournament and doesn't get paid what happens to the winner's share of the purse? The player(s) in second place the money divided between them, plus the money the receive for their finish.
Has anyone ever got a hole in one on every hole in a golf match? No, and I'm going to say no one ever will.
Who wins a match in raquetball? Chuck Norris! he always wins!
How soon will Black hole electron WikiAnswer be updated to match current Wikipedia Black hole electron article?
What is the most consecutive birdies made on the same hole by the same player in consecutive rounds by an amateur? 7
If a match in soccer is tied who wins? Nobody. It's a draw.
Who wins a football match if it's abandoned due to 5 people from 1 team been sent off? According to the FIFA Laws of the Game, a team must have 7 players for a match to continue. As such, after 5 players are sent off, a team of 11 has logically dropped to an unsustainable number and the match must be abandoned. The winner is not determined by the referee. There referee reports the number of goals scored for each team and the league or competition authority decides who the winner is based on this information. It is entirely possible that if both teams had an excessive amount of misconduct that neither team would receive the win. I have seen this happen in local leagues.
Does an amateur golfer win prize money E.g. The amateur that came joint fifth at the british open? Doesn't get a bean. Amateurs do not get any prize money.
What are you supposed to do when you are in a semis blind spot?
Do women have a hole behind there pee hole?
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