If You Flip a Coin and if You Get Heads You Flip Again What Are the Odds of the Second Being Heads
1) The mathematical theory of probability assumes that we have a well defined repeatable (in principle) experiment, which has equally its outcome a prepare of well divers, mutually exclusive, events.
Examples:
In the experiment of rolling one dice, the mutually exclusive outcomes are the dice landing with either the 1, two, iii, 4, five, or vi confront.
2) We assume that in whatever particular individual trial of the experiment, the outcome for that private trial cannot be predicted or known before paw - it is controlled by run a risk. Even so, when a very large number of contained trials of the experiment are performed, one finds that each possible result occurs a well defined fraction of the fourth dimension. This fraction is called the probability for that particular outcome to occur. The probability for an effect is always a number between 0 and ane. If the probability is zilch, we say that this outcome tin never occur. If the probability is one, we say that this result always occurs with complete certainty.
Examples:
When nosotros role a die a very large number of times, nosotros observe that nosotros get whatsoever given face up 1/half dozen of the time. The probability for the 1 face to appear is therefore 1/half-dozen. Similarly the probability for the 2 (3, 4, 5, or six) face to appear is also i/six.
iii) The probability for a given upshot might exist calculable from some underlying assumptions. Or the probability for an outcome can be determined experimentally by doing many trials.
Example:
When nosotros flip a money, we assume that there is no major difference between the 2 sides, and there is no exacting way we perform the flip in gild to try and influence the outcome. Because of all the random factors beyond our control that enter the flipping procedure (force with which the coin is flipped, motion of the air in the room, position of our paw when we take hold of the coin...) we therefore expect a probability of 1/two for heads, and 1/2 for tails. Each possible outcome is as likely. However, if we did a very large number of trial flips, and consistently found heads occurring 3/4 of the time, and tails 1/iv of the fourth dimension, we would know that our assumption of equally probable outcomes was false - we are dealing with a loaded money. Performing the experiment is the way to test the assumptions!
Allow's now come across how these ideas piece of work for some more complicated examples.
1) Consider the experiment of flipping of iv coins.
Each coin flip has ii possible outcomes, and then the flipping of four coins has iix2ten2x2 = xvi possible outcomes. We can enumerate all possible outcomes every bit follows, where H indicates a caput, and T a tail:
HHHH THHH
HHHT THHT
HHTH THTH
HHTT THTT
HTHH TTHH
HTHT TTHT
HTTH TTTH
HTTT TTTT
If nosotros assume that each individual coin is equally likely to come upwards heads or tails, then each of the above sixteen outcomes to 4 flips is as likely. Each occurs a fraction i out of 16 times, or each has a probability of 1/sixteen.
Alternatively, we could argue that the 1st coin has probability ane/2 to come up heads or tails, the 2nd money has probability ane/2 to come up heads or tails, so on for the 3rd and 4th coins, and then that the probability for any one particular sequence of heads and tails is simply (1/2)ten(one/2)x(1/2)10(1/2)=(1/16).
Now lets ask: what is the probability that in iv flips, ane gets North heads, where N=0, 1, ii, 3, or iv. We tin become this just by counting the number of outcomes above which accept the desired number of heads, and dividing by the total number of possible outcomes, 16.
N # outcomes with N heads probability to go N heads
0 1 1/16 = 0.0625
one 4 four/16 = ane/four = 0.25
ii 6 vi/16 = three/eight = 0.375
3 4 4/16 = 1/4 = 0.25
4 ane i/sixteen = 0.0625
We tin plot these results on a graph equally shown below.
The dashed line is shown just as a guide to the eye. Detect that the curve has a "bell" shape. The most likely outcome is for Northward=2 heads, where the curve reaches its maximum value. This is simply what you would expect: if each coin is as likely to land heads equally tails, in 4 flips, half should come up heads, that is N = fourten(1/two) = ii is the most likely outcome. Notation all the same that an occurrence of N = ane or Northward = 3 is not and so unlikely - they occur 1/four or 25% of the time. To have an occurrence of only Northward = 0, or Northward = 4 (no heads, or all heads) is much less likely - they occur but one/16 or half dozen.25% of the fourth dimension.
The in a higher place procedure is in principle the fashion to solve all problems in probability. Define the experiment, enumerate all possible mutually exclusive outcomes (which are usually assumed to be each equally likely), and so count the number of these outcomes which accept the particular property beingness tested for (here for example, the number of heads). Dividing this number by the full number of possible outcomes then gives the probability of the system to have that item property.
Often, however, the number of possible outcomes may be and then large that an explicit enumeration would go very tedious. In such cases, i tin resort to more subtle thinking to get in at the desired probabilities. For case, nosotros can deduce the probabilities to become N heads in iv flips as follows:
N=0: There is just one possible result that gives 0 heads, namely when each flip results in a tail. The probability is therefore i/16.
Northward=4: There is simply one possible outcome that gives 4 heads, namely when each flip results in a head. The probability is therefore 1/16.
N=ane: There are 4 possible outcomes which will have only one coin heads. Information technology may be that the 1st coin is heads, and all others are tails; or it may exist that the 2nd coin is heads, and all others are tails; or it may be that the third (or the quaternary) coin is heads, and all others are tails. Since there are 4 possible outcomes with one head but, the probability is 4/16 = 1/iv.
N=3: To get 3 heads, means that ane gets only one tail. This tail can be either the 1st coin, the 2nd coin, the 3rd, or the 4th coin. Thus there are merely four outcomes which take 3 heads. The probability is four/xvi = 1/4.
N=ii: To enumerate directly all the possible outcomes which have exactly 2 heads but, is a bit trickier than the other cases. We will come up to information technology shortly. But we tin can go the desired probability for N=2 the following way: Nosotros have already enumerated all possible outcomes with either N = 0, 1, 3, or 4 heads. These account for 1 + 4 + 4 + i = x possible outcomes. The only outcomes not include in these ten are those with exactly Due north=2 heads. Since there are sixteen possible outcomes, and 10 do non have N=ii heads, there must therefore be exactly 16 - 10 = 6 outcomes which do have exactly N=2 heads. The probability for N=2 is therefore 6/16 = 3/8.
two) Consider the experiment of rolling iii dice, each of which has half dozen sides.
What is the probability that no ii dice state with the same number side upwards, i.e. each of the three dice rolls a different number?
Since each die has six possible outcomes, the number of possible outcomes for the roll of three die is 6x6x6 = 216. We could enumerate all these 216 possibilities, and and so count the number of outcomes in which each die has a unlike number. This is clearly also tedious! Instead nosotros reason as follows:
When the 1st die is rolled, it can country with anyone of its vi faces upwardly. When the second dice is rolled, there are merely 5 possible sides which will be consequent with our benchmark that no two die land the same. When the third die is rolled, it must state with a face dissimilar from dice one and two; at that place are 4 possibilities. Therefore the total number of outcomes which have each of the three die with a dissimilar side upwardly is 6x5x4 = 120.
The probability for this to happen is and so 120/216 = v/nine = .55555...
Alternatively, we can say that the probability for the 1st die to state with any side up is one, the probability for the 2nd die to land with a side different from the 1st die is v/6, and the probability for the 3rd die to land with a side different from both the 1st and the 2nd die is 4/6. The probability that each of the three dice has a different side upward is then (1)x(5/six)x(4/6) = 20/36 = 5/9.
Nosotros larn from the above examples three general rules about probabilities:
1) The probability than whatever one of a given grouping of mutually exclusive possible outcomes volition actually occur in a given experiment, is the sum of the individual probabilities for each outcome in the grouping.
Example:
2) The probability for two independent events to both occur is just the production of the probabilities for each individual consequence.
Case:
Example:
1) If we roll 4 dice, what is the probability that at least i of them lands with the "6" face on acme?
Since each dice tin can land vi ways, and there are 4 die, there are half dozenxhalf dozenx6x6 = 64 = 1,296 dissimilar possible outcomes. Nosotros could in principle write these all down and so count the number which have at least one "6", just this is way too tedious. Instead, lets count the number of outcomes which do not accept whatsoever "6"south showing. To take no "6" showing, the first dice tin can land in any one of 5 possible ways. Similarly for the 2nd, 3rd, and 4th die. Thus the number of outcomes with no "6" showing on any die is 5x5x5x5 = 54 = 625. Since we must either run into no "6"south, or else we see at least ane "6", it follows that the number of outcomes with at least one "six" must be the (total number of outcomes) - (number of outcomes with no "vi") = 1,296 - 625 = 671. The probability to have at least one "6" is therefore 671/1296 = 0.5177469...
We could also have arranged our calculation as follows:
Written this mode, we see that (5/half-dozen) is the probability that any give die lands without a "6" on top. (5/vi)x(v/6)10(v/6)x(5/6) is therefore (general dominion #two) the probability that all four dice land without a "6" on top. Since the probability of all possible outcomes must sum to unity (full general rule #1), we therefore conclude that 1 - (five/6)x(5/vi)x(five/vi)x(5/6) is the probability that the opposite of "all 4 dice land without a "half-dozen" on top" happens. Just the reverse of "all 4 dice land without a "6" on top" is that "at least one dice lands with a "6" on superlative"!
In this example we see that we can split up all our possible outcomes into two classes. One class which has the desired holding we are testing for (in this case, at least one "6" on superlative), and a second class which does not take the desired property (no dice has a "6" on elevation). Often it turns out to be difficult to directly calculate the probability for the desired property, but like shooting fish in a barrel to calculate the probability not to find the desired property. The showtime probability is so just unity minus the second probability.
2) What is the probability that in a group of Due north people, nosotros will find at to the lowest degree two people with the same birthday? For the purpose of this question, we will consider people built-in on February 29 every bit if they were born on February 28!
This is equal to one - (probability no 2 people have same birthday)
Probability that no two people have the same birthday is:
since the first person can have whatever birthday, the second person can have simply one of the remaining 364 out of 365 possible birthdays (i.e. not the same every bit the first), the 3rd person can have but one of the remaining 363 out 365 possible birthdays (i.e. not the same as either the first or the 2d), and then on, down to the last Nth person. And then the probability that at least 2 people in a group of Due north have the same birthday is:
p(North) = one -
We can plot p(N) versus N on a graph, equally shown beneath.
Note that for N=10, p(x)=0.11695, or about 12% risk to have ii people with the same altogether. For North=l, p(l)=0.97037, or 97% take chances to accept two people with the aforementioned birthday. These probabilities are in full general much larger than most people's intuition would have guessed.
Examples from Text:
1) Two fair dice are rolled. What is the probability p that at least one die come upwards a 3?
We can practice this two ways:
i) The straightforward manner is as follows. To get at to the lowest degree one 3, would be consistent with the following three mutually exclusive outcomes:
the 1st dice is a iii and the 2nd is not: prob = (1/6)x(five/6)=5/36
the 1st die is not a iii and the 2nd is: prob = (5/6)x((1/6)=5/36
both the 1st and 2nd come up up 3: prob = (1/half-dozen)x(1/6)=1/36
sum of the in a higher place three cases is prob for at to the lowest degree one three, p = 11/36
ii) A faster way is as follows: prob at least one three = one - (prob no 3'southward)
The probability to get no 3's is (v/6)10(five/6) = 25/36.
So the probability to become at least one 3 is, p = 1 - (25/36) = 11/36
2) What is the probability that a card fatigued at random from an ordinary 52 deck of playing cards is a queen or a eye?
There are 4 queens and xiii hearts, and so the probability to draw a queen is
4/52 and the probability to draw a heart is xiii/52. But the probability to draw a queen or a heart is Non the sum four/52 + 13/52. This is because drawing a queen and drawing a eye are not mutually exclusive outcomes - the queen of hearts tin can meet both criteria! The number of cards which see the criteria of beingness either a queen or a eye is simply 16 - the 4 queens and the 12 remaining hearts which are non a queen. So the probability to depict a queen or a heart is xvi/52 = 4/13.
3) Five coins are tossed. What is the probability that the number of heads exceeds the number of tails?
We can divide all possible outcomes into the post-obit two mutually sectional groups:
i) the number of heads flipped is more than than the number of tails flipped
two) the number of tails flipped is more than the number of heads flipped
Since the probability to flip a caput is the same as the probability to flip a tail, the probability of outcome (i) must be equal to the probability of outcome (ii). So both must exist equal to 1/2.
Annotation that this answer works for whatever odd number of coin flips.
4) 4 boys and 3 girls are standing in a line. If the position of each child in line is random, what is the probability that the first 3 places in line are all girls?
We will solve this problem 2 means:
i) Lets us count the number of all the possible orderings of the seven children which are consequent with the desired outcome (i.e. 3 girls in front) and divide by the total number of all possible orderings.
The full number of orderings is: 7x6x5104x3x2ten1
as there are 7 children to pick to be 1st in line, only 6 remaining children to pick to be 2nd in line, then only 5 remaining children to selection to be third in line, and so on.
The number of orderings consistent with 3 girls in front is: (three10iitenane)x(4x3x2x1)
as at that place are 3 girls to option to be 1st in line, two remaining girls to selection to be second in line, and the last daughter must go 3rd in line; and so there are iv boys to place fourth in line, 3 remaining boys to place 5th in line, and so on.
So the probability to go all 3 girls in front is 
2) Alternatively, nosotros can argue as follows. The probability that a girl is 1st in line is (iii/seven) since 3 of the 7 children are girls. The probability that a daughter is 2d in line, given that the 1st is a daughter, is (2/6) since 2 of the remaining 6 children are girls. The probability that a girl is third in line, given that the 1st and 2nd are girls, is (1/5) since now but i of the remaining 5 children is a daughter. Then the probability that the 1st, 2nd, and third positions are all girls is
Source: http://www.pas.rochester.edu/~stte/phy104-F00/notes-2.html
0 Response to "If You Flip a Coin and if You Get Heads You Flip Again What Are the Odds of the Second Being Heads"
Post a Comment