Algebra

Probabilities

If all the ways in which an event can happen be 𝑚 in number, all being equally likely to occur, and if in 𝑛 of these 𝑚 ways the event would happen under certain restrictive conditions; then the probability of the restricted event happening is equal to 𝑛÷𝑚.
Thus, if the eltters of the alphabet be chosen at random, any letter being equally likely to be taken, the probability of a vowel being selected is equal to 526. The number of unrestricted cases here is 26, and the number of restricted ones 5. 309 If, however, all the 𝑚 events are not equally probable, they may be divided into groups of equally probable cases. The probability of the restricted event happening in each group separately must be calculated, and the sum of these probabilities will be the total probability of the restricted event happening at all.

Example

There are three bags 𝐴, 𝐵, and 𝐶. 𝐴 contains 2 white and 3 black balls 𝐵 contains 3 white and 4 black balls 𝐶 contains 4 white and 5 black balls A bag is taken at random and a ball drawn from it. Required the probability of the ball being white. Here the probability of the bag 𝐴 being chosen =13, and the subsequent probability of a white ball being drawn =25. Therefore the probability of a white ball being drawn from 𝐴 =13×25=215 Similarly the probability of a white ball being drawn from 𝐵 =13×37=17 And the probability of a white ball being drawn from 𝐶 =13×49=427 Therefore the total probability of a white ball being drawn =215+17+427=401945 310 If 𝑎 be the number of ways in which an event can happen, and 𝑏 the number of ways in which it can fail, then the Probability of the event happening = 𝑎𝑎+𝑏 311 Probability of the event failing =𝑏𝑎+𝑏 Thus Certainty =1 312 If 𝑝, 𝑝' be the respective probabilities of two independent events, then Probability of both happening =𝑝𝑝' 313 Probability of not both happening =1−𝑝𝑝' 314 Probability of one happening and one failing=𝑝+𝑝'−2𝑝𝑝' 315 Probability of both failing =(1−𝑝)(1−𝑝') 316 If the probability of an event happening in one trial be 𝑝, and the probability of its failing 𝑞, then Probability of the event happening 𝑟 times in 𝑛 trials =𝐶(𝑛,𝑟)𝑝^𝑟𝑞^𝑛−𝑟 317 Probability of the event failing 𝑟 times in 𝑛 trials =𝐶(𝑛,𝑟)𝑝^𝑛−𝑟𝑞^𝑟 By induction 318 Probability of the event happening at least 𝑟 times in 𝑛 trials= the sum of the first 𝑛−𝑟+1 terms in the expansion of (𝑝+𝑞)^𝑛. 319 Probability of the event failing at least 𝑟 times in 𝑛 trials= the sum of the last 𝑛−𝑟+1 terms in the expansion. 320 The number of trials in which the probability of the same event happening amounts to 𝑝' =log(1−𝑝')log(1−𝑝) From the equation (1−𝑝)^𝑥=1−𝑝' 321

Definition:

When a sum of money is to be received if a certain event happens, that sum multiplied into the probability of the event is termed the expectation.

Example

If three coins be taken at random from a bag containg one sovereign, four half-crowns, and five shillings, the expectation will be the sum of the expectations founded upon each way of drawing three coins. But this is also equal to the average value of three coins out of the ten; that is, 310ths of 35 shillings, or 10𝑠. 6𝑑.322 The probability that, after 𝑟 chance selections of the numbers 0, 1, 2, 3, ⋯, 𝑛, the sume of the numbers drawn will be 𝑠, is equal to the coefficient of 𝑥^𝑠 in the expansion of (𝑥⁰+𝑥¹+𝑥²+⋯+𝑥^𝑛)^𝑟÷(𝑛+1)^𝑟 323 The probability of the existence of a certain cause of an observed event out of several known causes, one of which must have produced the event, is proportional to the a priori probability of the cause existing multiplied by the probability of the event happening from it if it does exist.
Thus, if the a priori probabilities of the causes be 𝑃₁, 𝑃₂, ⋯, and the corresponding probabilities of the event happening from those causes 𝑄₁, 𝑄₂, ⋯, then the probability of the 𝑟^th cause having produced the event is 𝑃_𝑟𝑄_𝑟∑(𝑃𝑄) 324 If 𝑃′₁, 𝑃′₂, ⋯, be the a priori probabilities of a second event happening from the same causes respectively, then, after the first event has happened, the probability of the second happening is ∑(𝑃𝑄𝑃')∑(𝑃𝑄) For this is the sum of such probabilities as ∑(𝑃_𝑟𝑄_𝑟𝑃′_𝑟)∑(𝑃𝑄), which is the probability of the 𝑟^th cause existing multiplied by the probability of the second event happening from it.

Example 1

Suppose there are 4 vases containing each 5 white and 6 black balls. 2 vases containing each 3 white and 5 black balls and 1 vase containing 2 white and 1 black ball. A white ball has been drawn, and the probability that it came from the group of 2 vases is required. Here 𝑃₁=47𝑃₂=27𝑃₃=17 𝑄₁=511𝑄₂=38𝑄₃=23 Therefore, by (324), the probability required is 2⋅37⋅84⋅57⋅11+2⋅37⋅8+1⋅27⋅3=99427

Example 2

After the white ball has been drawn and replaced, a ball is drawn again; required the probability of the ball being black. Here 𝑃′₁=611, 𝑃′₂=58, 𝑃′₃=13, The probability, by (325), will be 4⋅5⋅67⋅11⋅11+2⋅3⋅57⋅8⋅8+1⋅2⋅17⋅3⋅34⋅57⋅11+2⋅37⋅8+1⋅27⋅3=58639112728 If the probability of the second ball being white is required, 𝑄₁𝑄₂𝑄₃ must be employed instead of 𝑃′₁𝑃′₂𝑃′₃. 325 The probability of one event at least happening out of a number of event whose respective probabilities are 𝑎, 𝑏, 𝑐, ⋯ is 𝑃₁−𝑃₂+𝑃₃−𝑃₄+⋯ where 𝑃₁ is the probability of 1 event happening, 𝑃₂ is the probability of 2 event happening, and so on. For, by (316), the probability is 1−(1−𝑎)(1−𝑏)(1−𝑐)⋯=∑𝑎−∑𝑎𝑏+∑𝑎𝑏𝑐−⋯ 326 The probability of the occurrence of 𝑟 assigned events and no more out of 𝑛 events is 𝑄_𝑟−𝑄_𝑟+1+𝑄_𝑟+2−𝑄_𝑟+3+⋯ where 𝑄_𝑟 is the probability of the 𝑟 assigned events; 𝑄_𝑟+1 the probability of 𝑟+1 events including the 𝑟 assigned events For if 𝑎, 𝑏, 𝑐, ⋯ be the probabilities of the 𝑟 events, and 𝑎', 𝑏', 𝑐', ⋯ be the probabilities of the excluded events, the required probability will be 𝑎𝑏𝑐⋯(1−𝑎')(1−𝑏')(1−𝑐')⋯ =𝑎𝑏𝑐⋯(1−∑𝑎'+∑𝑎'𝑏'−∑𝑎'𝑏'𝑐'+⋯ 327 The probability of any 𝑟 events happening and no more is ∑𝑄_𝑟−∑𝑄_𝑟+1+∑𝑄_𝑟+2−⋯ Note : If 𝑎=𝑏=𝑐=⋯, then ∑𝑄_𝑟=𝐶(𝑛,𝑟)𝑄_𝑟, ⋯ 328

Sources and References

https://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdrive