Introduction on Probability Simple Events Introduction:
The probability is possible simple events is divided into total number of simple events this is called the probability. The probability are contains two types of distributions. These are the discrete distribution and continuous distribution. The general formation of the probability
No. of possible events(x)
The probability of event A = ----------------------------------
Total no. of events(n)
``
Probability Simple Event - Example 1:
Arnold and John play a fair game repeatedly for one nickel each game. If originally Arnold has a nickels and John has b nickels, what is Arnold's chances of winning all of John's money, assuming the play goes on until one person has lost all her or his money?
Simple Event Solution:
Let p (n) be Arnold's chances of winning the total amount of a + b, provided she has n nickels in her possession. Obviously p(0) = 0. If she is left with a non-zero capital, Arnold may, at every trial, win or lose one nickel, both with the probability of 1/2,
P (n) =` (p (n+1))/2 + (p (n-1))/2` , n > 0
In other words, 2 p (n) = p (n + 1) + p (n - 1), or p (n + 1) – p (n) = p (n) – p (n - 1). From here, recursively,
p(n + 1) - p(n) = p(n) - p(n - 1)
= p (n - 1) – p (n - 2)
= p (n - 2) – p (n - 3)….p (2) – p (1)
= p (1)-p (0)
= p (1)
It follows that p(n) = n p(1) and, since, p(a + b) = 1, p(1) = 1 / (a + b). It follows that p(a) = a / (a + b).
Between, if you have problem on these topics Adding Percentages, please browse expert math related websites for more help on Scientific Notation Rules.
Probability Simple Event- Example 2:
An urn contains w orange balls and b blue balls (w > 0 and b > 0). The balls are mixed and two balls are drawn from the urn, one after the other, without replacement. Let Wi and Bi denote the respective outcomes 'orange on the ith draw' and 'blue on the ith draw,' for i = 1, 2.
Solution:
Prove that P(W2) = P(W1) = w/(w + b). (Which clearly implies a similar identity for B2 and B1?)
Furthermore, P (Wi) = w/ (w + b), for any i not exceeding the total number of balls w + b.
We must remember the formula for the total probability:
P (W2) = P(W2|W1)·P(W1) + P(W2|B1)·P(B1),
From which
P(W2) = (w - 1)/(b + w - 1)·w/(b + w) + w/(b + w - 1)·b/(b + w)
= w/(b + w - 1)(b + w)·(w - 1 + b),
Which is simplified to w/(w + b).
My Previous Blog :- http://onlinemathsolver.blogspot.in/2012/10/definition-permutation.html
The probability is possible simple events is divided into total number of simple events this is called the probability. The probability are contains two types of distributions. These are the discrete distribution and continuous distribution. The general formation of the probability
No. of possible events(x)
The probability of event A = ----------------------------------
Total no. of events(n)
``
Probability Simple Event - Example 1:
Arnold and John play a fair game repeatedly for one nickel each game. If originally Arnold has a nickels and John has b nickels, what is Arnold's chances of winning all of John's money, assuming the play goes on until one person has lost all her or his money?
Simple Event Solution:
Let p (n) be Arnold's chances of winning the total amount of a + b, provided she has n nickels in her possession. Obviously p(0) = 0. If she is left with a non-zero capital, Arnold may, at every trial, win or lose one nickel, both with the probability of 1/2,
P (n) =` (p (n+1))/2 + (p (n-1))/2` , n > 0
In other words, 2 p (n) = p (n + 1) + p (n - 1), or p (n + 1) – p (n) = p (n) – p (n - 1). From here, recursively,
p(n + 1) - p(n) = p(n) - p(n - 1)
= p (n - 1) – p (n - 2)
= p (n - 2) – p (n - 3)….p (2) – p (1)
= p (1)-p (0)
= p (1)
It follows that p(n) = n p(1) and, since, p(a + b) = 1, p(1) = 1 / (a + b). It follows that p(a) = a / (a + b).
Between, if you have problem on these topics Adding Percentages, please browse expert math related websites for more help on Scientific Notation Rules.
Probability Simple Event- Example 2:
An urn contains w orange balls and b blue balls (w > 0 and b > 0). The balls are mixed and two balls are drawn from the urn, one after the other, without replacement. Let Wi and Bi denote the respective outcomes 'orange on the ith draw' and 'blue on the ith draw,' for i = 1, 2.
Solution:
Prove that P(W2) = P(W1) = w/(w + b). (Which clearly implies a similar identity for B2 and B1?)
Furthermore, P (Wi) = w/ (w + b), for any i not exceeding the total number of balls w + b.
We must remember the formula for the total probability:
P (W2) = P(W2|W1)·P(W1) + P(W2|B1)·P(B1),
From which
P(W2) = (w - 1)/(b + w - 1)·w/(b + w) + w/(b + w - 1)·b/(b + w)
= w/(b + w - 1)(b + w)·(w - 1 + b),
Which is simplified to w/(w + b).
My Previous Blog :- http://onlinemathsolver.blogspot.in/2012/10/definition-permutation.html
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