Electron Probability Distribution

Introduction to electron probability distribution:

The probability distribution is the major part in the probability theory and the statistics. The probability distribution is used to determine the number of possibility for the occurrence of an event. The most commonly used probability distributions are the binomial distribution, geometric distribution, normal distribution and the gamma distribution. These above mentioned distributions are included in the discrete and continuous probability distribution. The major type of the probability distribution is the discrete probability distribution and the continuous probability distribution. This article has the study about electron probability distribution.

Types of Electron Probability Distribution:

The major types of the probability distribution are

Discrete probability distribution

Continuous probability distribution

Discrete probability distribution:

The probability for a countable number of occurrences for the event is calculated in the discrete probability distribution.

Continuous probability distribution:

The probability values in this are the continuous ranged value it is calculated in the continuous probability distribution.

Examples for Electron Probability Distribution:

Example 1 to electron probability distribution:

A manufacturer of cotton pins knows that 7 % of his product is defective. If he sells pins in boxes of 100 and guarantees that not more than 4 pins will be defective. Determine the probability for a box will fail to meet the guaranteed quality.

Solution:

The value of p is p = '7/100' , n = 100

The mean value is 'lambda' = n p = ('7/100' ) (100) = 7

By the Poisson distribution

P[X = x] = '(e^ - lambda lambda^x)/ (x!)'

Probability for a box will to meet the guaranteed quality = P[X > 4]

P[X > 4] = 1- P[X = 4]

P[X > 4] = 1- (P (0) +P (1) +P (2) +P (3) +P (4))

P[X > 4] = 1- 'e^-7' (1 + 7 + '49/2' + '343/6' + '2401/ 24' )

P[X > 4] = 1- 'e^-7' (1+ 7 + 24.5 + 57.17 + 100.04)

P[X > 4] = 1- 'e^-7' (189.71)

P[X > 4] = 1- 0.00091(189.71)

P[X > 4] = 1- 0.1726

P[X > 4] = 0.8274

The probability for the box will fail to meet the guaranteed quality is 0.8274.

Example 2 to electron probability distribution:

The probability for destroying the target in only one time is 0.40. Compute the probability that it would be destroyed on the third attempt itself.

Solution:

The probability of destroying the target in one trial is p = 0.40

The value of the q is calculated by q = 1-p

q = 1- 0.40

q = 0.60

By the geometric distribution, the probability for the success is calculated by using the formula

P(X =x) = q x p, the value of x is 0, 1, 2. . .

The target is destroyed at the third attempt, so x = 3.

P(X = 3) = (0. 60) 3 (0.40)

P(X = 3) = (0.216) (0.40)

P(X = 3) = 0.0864

The probability for destroying the target at the third trial is 0.0864.

by:Omkar Nayak

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