Probability, Permutations and Combinations

[bella1234]

<p>This is my weak area and I could use some help. Can someone point out book/resource/strategy that might have helped you hone up this area?</p>

[anishbasu]

<p>Probability is really easy when you master the algorithm:
For example:
Imagine a reality show. This is the final round. The host has asked the final contestant that he must choose from the set of 5 doors, which one has the jackpot of $100,000.
Now, By the general definition of probability:

Probability P(E) = Number of favourable events n(e)/Total Events n(s)

Number of favourable events n(e) = 1 (Since the jackpot is behind only 1 door)
Total number of events n(s) = 5 (ie. The total number of doors, or chances)
Now, if the contestant chooses a door(Say, the 3rd), the Probability of it having the jackpot:

P(E) = n(e)/n(s)
or, P(e) = 1/5
or, P(e) = 0.20

Now, the host reveals the 1st door, there are no prizes: (n(s) reduced by 1 due to open door)</p>

<p>Number of favourable events n(e) = 1 (Since the jackpot is behind only 1 door)
Total number of events n(s) = 4 (ie. The total number of doors, or chances)

P(E) = n(e)/n(s)
or, P(e) = 1/4
or, P(e) = 0.25

Hmmm, a better chance at winning this.</p>

<p>Host reveals the 5th Door: (No prizes here)
Number of favourable events n(e) = 1 (Since the jackpot is behind only 1 door)
Total number of events n(s) = 3 (ie. 2 Doors are open(5-2=3))

P(E) = n(e)/n(s)
P(E) = 1/3
P(E) = 0.333333333(Better Chance!)

The 2nd Door is Opened(No prizes)
Number of favourable events n(e) = 1 (Since the jackpot is behind only 1 door)
Total number of events n(s) = 2 (ie. 2 Doors are open(5-3=2))
2 Doors left 4 and 3.

P(E) = n(e)/n(s)
P(E) = 1/2
P(E) = 0.5(Even better chance!)

The third door is Opened:
** The Jackpot is in the 3rd Door! **
So, the contestant chose the correct Door a 1 in 5 chance or a Probability of 0.20.</p>

<p>Now that our contestant is successful, we can explore the chances of him being unsuccessful as:

P(E)’ = 1-P(E)

.`. P(E)’ = 1 - 1/5
=> P(E)’ = 4/5
=> P(E)’ = 0.80

Or, he had 0.8 Probability of not winning the game.</p>

<p>Probability is just like percentages. It is the winning moves you can make, divided by the total number of moves</p>

[anishbasu]

<p>Permutations and combinations:
[Easy</a> Permutations and Combinations | BetterExplained](<a href=“http://betterexplained.com/articles/easy-permutations-and-combinations/]Easy”>Easy Permutations and Combinations – BetterExplained)</p>

[bella1234]

<p>Thanks Guys, appreciate the time.</p>