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Probability in everyday life

Probability is connected with uncertainty. In any statistical experiment, the outcomes that occur may be known, but exactly which one might occur is usually not known. Mathematically, probability theory formulates incomplete knowledge pertaining to the likelihood of an occurrence. For example, a meteorologist might say there is a 60% chance that it will rain tomorrow. This means that in 6 of every 10 times when the world is in the current state, it will rain tomorrow.

In everyday speech, we often refer to probabilities using percentages between 0% and 100%. A probability of 100% (100 out of 100) means that an event is certain, whereas a probability of 0% (0 out of 100) is often taken to mean the event is impossible. However, in certain circumstances, there can be a distinction between logically impossible and occurring with zero probability if the sample space is large enough (i.e. infinite) or in some other way indeterminate. For example, in selecting a random fraction between 0 and 1, the probability of selecting 1/2 is 0, but it is not logically impossible (since there are infinitely many numbers), while trying to predict the exact number of raindrops to fall into a large enough area is also 0 (or close to it), since the possible number of raindrops that might occur at a given time is both very large and difficult to determine.

Another way of referring to probabilities is odds. The odds of an event is defined as the ratio of the probability that the event occurs to the probability that it does not occur. For example, the odds of a coin landing on a given side are 0 . 5 0 . 5 = 1 , usually written "1 to 1" or "1:1". This means that on average, the coin will land on that side as many times as it will land on the other side.

The simplest example: equally likely outcomes

We say two outcomes are equally likely if they have an equal chance of happening. For example when a fair coin is tossed, each outcome in the sample space S = { h e a d s , t a i l s } is equally likely to occur.

Probability is a function of events (since it is not possible to have a single event with two different probabilities occurring), so we usually denote the probability P of some event E occurring by P ( E ) . When all the outcomes are equally likely (in any activity), it is fairly straightforward to count the probability of a certain event occuring. In this case,

P ( E ) = n ( E ) / n ( S )

For example, when you throw a fair die the sample space is S = { 1 ; 2 ; 3 ; 4 ; 5 ; 6 } so the total number of possible outcomes n ( S ) = 6 .

Event 1: The value of the topmost face of the die (call it 'k') is 4

The only possible outcome is a 4 , i.e E = { 4 } . So n ( E ) = 1 .

Probability of getting a 4: P ( k = 4 ) = n ( E ) / n ( S ) = 1 / 6 .

Event 2: The value of the topmost face is a number greater than 3

Favourable outcomes: E = { 4 ; 5 ; 6 }

Number of favourable outcomes: n ( E ) = 3 .

Probability of getting a number greater than 3: P ( k > 3 ) = n ( E ) / n ( S ) = 3 / 6 = 1 / 2 .

A standard deck of cards (without jokers) has 52 cards. There are four sets of cards, called suits. The suit a card belongs to is denoted by a symbol on the card, the four possible symbols being hearts, clubs, spades, and diamonds. In each suit there are 13 cards ( 4 suits × 13 cards = 52 ) consisting of one each of ace, king, queen, jack, and the numbers 2-10.

If we randomly draw a card from the deck, we can the card drawn as a possible outcome. Therefore, there are 52 possible outcomes. We can now look at various events and calculate their probabilities:

  1. Out of the 52 cards, there are 13 clubs. Therefore, if the event of interest is drawing a club, there are 13 favourable outcomes, what is the probability of this event?
  2. There are 4 kings (one of each suit). The probability of drawing a king is?
  3. What is the probability of drawing a king OR a club?
  1. The probability of this event is 13 52 = 1 4 .

  2. 4 52 = 1 13 .

  3. This example is slightly more complicated. We cannot simply add together the number of number of outcomes for each event separately (4 + 13 = 17) as this inadvertently counts one of the outcomes twice (the king of clubs). Why is this so? Well, as was noted in question 3, part (f) above, n ( A B ) = n ( A ) + n ( B ) - n ( A B ) . In the results involving the throwing of dice, the intersection of any two outcomes was empty (and hence n ( A B ) = 0 ) since it is not possible for the top face of a die to have two different values simultaneously. However, in this case, a card can be both a club and a king at the same time (i.e. n ( A B ) = 1 ). Therefore, n ( A B ) = 4 + 13 - 1 = 16 . So the correct answer is 16 52 .

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Source:  OpenStax, Siyavula textbooks: grade 10 maths [ncs]. OpenStax CNX. Aug 05, 2011 Download for free at http://cnx.org/content/col11239/1.2
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