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occurs, in a large number of cases, to the whole number of times in which it occurs and in which it fails.

400. If an event is certain it will occur without fail in every case its probability is therefore unity.

It follows at once from the definition of probability that if p be the probability that any event should occur, p will be the probability of its failing to occur.

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When the probability of the happening of an event is to the probability of its failure as a is to b, the odds are said to be a to b for the event, or b to a against it, according as a is greater or less than b.

401. Exclusive events. Events are said to be mutually exclusive when the supposition that any one takes place is incompatible with the supposition that any other takes place.

When different events are mutually exclusive the chance that one or other of the different events occurs is the sum of the chances of the separate events.

It will be sufficient to consider three events.

Let the respective probabilities of the three events, expressed as fractions with the same denominator, be

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Then, out of d equally likely ways, the three events can happen in a, a, and a ways respectively.

Hence, as the events never concur, one or other of them will happen in a + a + as out of d equally likely Hence the probability of one or other of the three events happening is

ways.

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This proves the proposition for three mutually exclusive events; and any other case can be proved in a

similar manner.

Ex. 1. Find the chance of throwing 3 with an ordinary six-faced die.

Since any one face is as likely to be exposed as any other face, there is one favourable and five unfavourable cases which are all equally likely; hence the required probability is

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Ex. 2. Find the chance of throwing an odd number with an ordinary

die.

Ex. 3.

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Find the chance of drawing a red ball from a bag which contains 5 white and 7 red balls.

Here any one ball is as likely to be drawn as any other; thus there are 7 favourable and 5 unfavourable cases which are all equally 7 likely; the required probability is therefore 12°

Ex. 4. Two balls are to be drawn from a bag containing 5 red and 7 white balls; find the chance that they will both be white.

Here any one pair of balls is as likely to be drawn as any other pair. The total number of pairs is 12C2, and the number of pairs which are both white is C: the required chance is therefore

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Ex. 5. Shew that the odds are 7 to 3 against drawing 2 red balls from a bag containing 3 red and 2 white balls.

Ex. 6. Three balls are to be drawn from a bag containing 2 black, white and 2 red balls; shew that the odds are 3 to 2 against drawing a ball of each colour, and 4 to 1 against drawing 2 white balls.

Ex. 7. A party of n persons take their seats at random at a round table: shew that it is n-3 to 2 against two specified persons sitting together.

402. Independent Events. The probability that two independent events should both happen is the product of the separate probabilities of their happening.

Suppose that the first event can happen in a, and fail in b, equally likely ways; and suppose that the second event can happen in a, and fail in b, equally likely ways. Then each of the a, b, cases may be associated with each of the a, b, cases to make (a + b) (α, + b) compound cases which are all equally likely; and in aa, of these compound cases both events happen. Hence the proba

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Thus the probability of the concurrence of two independent events whose respective probabilities are p1 and p2 is P1 × P2

Cor. If Pi and P2 be the probabilities of two independent events, the chance that they will both fail is (1 - p) (1 - p), the chance that the first happens and the second fails is p, (1 - p), and the chance that the second happens and the first fails is (1 - P1) P2

It can be shewn in a similar manner that, if p1, P2, P3....... be the probabilities of any number of independent events, then the probability that they all happen will be p1· P2 ⋅ P3· · ·, and that they all fail (1 - p) (1 - P2) (1 - p)..., &c.

403. Dependent Events. If two events are not independent, but the probability of the second is different when the first happens from what it is when the first fails, the reasoning of the previous article will still hold good provided that p, is the probability that the second event happens when the first is known to have happened. Thus if P1 be the probability of any event, and p, the probability of any other event on the supposition that the first has happened; then the probability that both events will happen will be p1 × P2 And similarly for any number of dependent events.

Ex. 1. Find the probability of throwing two heads with two throws of a coin.

1

The probability of throwing heads is for each throw; hence the

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2

1 1

2

Ex. 2. Find the probability of throwing one 6 at least in six throws with a die.

The probability of not throwing 6 is

6

in each throw. Hence the

probability of not throwing a 6 in six throws is, by Art. 402,

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and therefore the probability of throwing one six at least is

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Ex. 3. Find the chance of drawing 2 white balls in succession from a bag containing 5 red and 7 white balls, the balls drawn not being replaced.

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The chance of drawing a white ball the first time is ; and, 12 having drawn a white ball the first time, there will be 5 red and 6 white balls left, and therefore the chance of drawing a white ball the second time will be Hence, from Art. 403, the chance of

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Ex. 4. There are two bags, one of which contains 5 red and 7 white balls and the other 3 red and 12 white balls, and a ball is to be drawn from one or other of the two bags; find the chance of drawing a red ball.

The chance of choosing the first bag is

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and if the first bag be

chosen the chance of drawing a red ball from it is

5

12; hence the 1 5 5 chance of drawing a red ball from the first bag is X =

2 12 24

Similarly the chance of drawing a red ball from the second bag is

1 3 1

X =

2 15 10'

Hence, as these events are mutually exclusive, the

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Ex. 5. In two bags there are to be put altogether 2 red and 10 white balls, neither bag being empty. How must the balls be divided so as to give to a person who draws one ball from either bag, (1) the least chance and (2) the greatest chance of drawing a red ball.

[The least chance is when one bag contains only one white ball, and the greatest chance is when one bag contains only one red ball,

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404. When the probability of the happening of an event in one trial is known, the probability of its happening exactly once, twice, three times, &c. in n trials can be at once written down.

For, if p be the probability of the happening of the event, the probability of its failing is 1 - p = q. Hence, from Art. 402, the probability of its happening r times and failing nr times in any specified order is p"q"-". But the whole number of ways in which the event could happen r times exactly in n trials is C, and these ways are all equally probable and are mutually exclusive. Hence the probability of the event happening r times exactly in n trials is C.p"q".

n

"

Thus, if (p + q)" be expanded by the binomial theorem, the successive terms will be the probability of the happening of the event exactly n times, n − 1 times, n − 2 times, &c. in n trials.

Cor. I. To find the most probable number of successes and failures in n trials it is only necessary to find the greatest term in the expansion of (p + q)”.

Cor. II. The probability of the event happening at least r times in n trials is

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Ex. 1. Find the chance of throwing 10 with 4 dice.

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The whole number of different throws is 64, for any one of six numbers can be exposed on each die; also the number of ways of throwing 10 is the coefficient of x1o in (x + x2 + +6), for this coefficient gives the number of ways in which 10 can be made up by the addition of four of the numbers 1, 2, 6, repetitions being allowed. Now the coefficient of x10 in (x+x2+ +x6)4, that is in x1 is easily found to be 80. Hence the required chance is

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-x64

-x

5

Ex. 2. Find the chance of throwing 8 with two dice.

Ans.

36

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