On the 5th November 1854 a smaller British and (yes) French army beat off an assault by the Russians at Inkerman in the Crimea. It was known as “The Soldier’s Battle” as men fought small engagements due to poor visibility in dense fog.
The Russians had massed 32000 men on the Allied flank and headed for the 2700 man 2nd Division, commanded today by the aggressive Pennefather. Instead of falling back in the face of superior numbers, he advanced. The British had their rifles to thank this day as they took a terrible toll on the musket armed Russian Infantry, who were hemmed in by the valley’s bottle neck shape. The British 2nd Division pushed the Russians back onto their reinforcements and should have been routed by the Russians’ numbers, but the fog and the British Light Division saved them. Three successive Russian commanders were killed in this engagement.
The Russians other 15000 men approached and assailed the Sandbag Battery, but they were routed by 300 British defenders vaulting the wall, blunting the lead Battalions, who were then attacked in the flank. More Russian attacks ensured the Battery exchanged hands several times.
The British 4th Division was not as lucky. Arriving on the field, its flanking move was itself flanked and its commander, Cathcart, killed. This enabled the Russians to advance, but not for long. They were soon driven off by French units arriving from their camps and made no more headway.
The battle was lost and they had to withdraw.
This was the last time the Russians tried to defeat the Allied troops in the field. Despite this reverse, however, the Russian attack had seriously stalled the Allies from capturing Sevastopol. They had to instead, spend one harsh winter on the heights overlooking the city, before it fell in September of 1855.
The British suffered 2573 casualties, the French 1800 and the Russians 11959.
Math help
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you have a hat with ten “numbers”. there are 3 0s, 3 2s and 4 1s. you have to get a sum of 3. you get to draw two numbers out. you put the 1st number back in and then draw the second( so results are indpentant) what is the chance of you getting that sum of 3 or better?
i really want to know how to do this so i can use it for a differnt probability problem.
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math is definently not my strong point
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Ok, first of all, how can you get a sum of 3? You need to draw 1 “one” and 1 “two” right? You can’t draw 2 of the same. You can’t draw any "zero"s. So if you know you must draw a “one” and a “two” and you can draw them in any order. So successful outcomes are as follows:
FIRST you draw a “one” THEN you draw a “two”. or
FIRST you draw a “two” THEN you draw a “one”.p1 is 4/10. p2 is 3/10. Then p1 and p2 = 12/100. Also p2 and p1 = 12/100
so add them together and your probability is 24/100. or 6/25.
Maybe I made a mistake? Anyone else?
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I think it’s easier to think of all the combinations.
There are 8 possible ways to draw numbers. Just write them out. Then look to see which ones equal three or more. Out of all the possibilities, only three will get you the desired sum ( 1/2, 2/1, 2/2).
But the likelihood of drawing a one is better than drawing a 0 or 2.
So let’s see what happens if we draw 1 first. That’s 4/10 probability. We MUST draw a 2, which is a 3/10 probability. That is 12/100, or 3/25, overall probability of happening (multiply the two probabilities).
Let’s see what happens if we draw 2 first. That’s a 3/10 probability. You can draw a 2 or 1, which is a 7/10 probability. That makes a 21/100 probability overall.
Since either can happen, we find that you have a 33/100 probability of achieving the desired amount, or pretty much 1/3…
(Sorry, rj - if I am right, and it has been a LONG time, then you are my inspiration…)
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Jermo,
I think you are correct if Cyan needs a sum of 3 or more:
@cyan:what is the chance of you getting that sum of 3 or better?
I was basing my solution on my understanding that he needed a sum of exactly 3:
@cyan:you have to get a sum of 3
Since it’s kind of ambiguous in the question, Cyan will need to clear that up. Depending on the question, one of these solutions should satisfy him :wink:
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Jermo,
I think you are correct if Cyan needs a sum of 3 or more:
@cyan:what is the chance of you getting that sum of 3 or better?
I was basing my solution on my understanding that he needed a sum of exactly 3:
@cyan:you have to get a sum of 3
Since it’s kind of ambiguous in the question, Cyan will need to clear that up. Depending on the question, one of these solutions should satisfy him :wink:
sorry i meant 3 or better. but i was hoping for an equation or something to make it easier. like for premutations or combunations. thanks for the help, i tried what jermo said and it worked…
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Well, as a combination I guess it is:
[(4 choose 1) x (3 choose 1)] + [(3 choose 1) x (7 choose 1)]but since you’re only choosing 1 each time, there really isn’t a need to use combinations / permutations.
Cheers.
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I admit I didn’t know any shortcuts - I was just trying to break it down better so cyan could see the meat & potatoes of the thing…
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you have a hat with ten “numbers”. there are 3 0s, 3 2s and 4 1s. you have to get a sum of 3. you get to draw two numbers out. you put the 1st number back in and then draw the second( so results are indpentant) what is the chance of you getting that sum of 3 or better?
i really want to know how to do this so i can use it for a differnt probability problem.
33%
You could just draw the matrix of 10x10 cells and label all the possible combinations then count the ones that work.
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AN EQUATION!!! :-o
Next thing you know you will be discussing the polarity of ZERO!!! :-P
You defined 10 possible selections so the odds of any single selection is 1/10.
You defined the possible selections as three 0’s, four 1’s and three 2’s.
This means the odds of selecting a 0 is 3/10 or 30% in a single draw. For a 1 it is 4/10 or 40% and for a 2 it is 3/10 or 30%.
You then specify two draws with replacement between the draws.
Finally, you requested the sum of the two draws equal or exceed “3”.
The possible results of the two draws are (x,y) where x is the first draw and y is the second draw.
These possible results are (0,0); (0,1); (0,2); (1,0); (1,1); (1,2); (2,0); (2,1); (2,2).
The only results you are interested in are (1,2); (2,1) and (2,2).
These are (0.4 * 0.3) + (0.3 * 0.4) + (0.3 *0.3)
This equals 0.12 + 0.12 + 0.09 or 0.33 which is 33%.
For a bunch of fancy pants math terms take a look at this:
http://en.wikipedia.org/wiki/Dice#Probability