The geometric distribution can be used for an experiment that has only 2 outcomes and is
repeated until a successful outcome occurs. For instance spinning a roulette wheel until a 7
is obtained, or rolling a die until a 5 occurs. In these situations a success would occur on the
nth trial, and a geometric distribution tells us the probability of when the success is likely to
happen. The geometric distribution satisfies the following conditions. Number 1 is that
each trial has only 2 outcomes, a success, or a failure. Number 2 is that each outcome is
independent of one another. Number 3 is the probability of success is the same for
each trial, and number 4 is that the experiment will continue until a success occurs. The formula
for the geometric distribution is probability of n equals, p times the quantity, 1 minus p raised to
the n minus 1. All right, let's break this formula down, probability of n means the probability that
the first success will occur on the nth trial. For instance the probability of getting the
first 5 will occur in the third roll of a die, so probability of n would be probability of
3, for the third die roll. Small p in the equation represents the given probability. In our die roll
the probability of a 5, is 1 over 6, as there are 6 possible numbers or outcomes. And n in
the formula is the number of trials the first success occurs on. This will be the same number
is our probability number, so again, the probability that the first 5 will occur on the third roll of a
die. So roll 1 is a failure, roll 2 is a failure, and roll 3 is a success, so n equals 3 because
this is the number of trials we're looking for the first success on. One note, this formula is
sometimes written as probability of x, equals, small p, times q, raised to the x minus 1, and in
this formula, x represents the number of trials the first success is on, and the q is equal to 1
minus p, which we have used in other distributions, but these formulas are the same. Let's look at a
couple of examples. First we'll stick with the die roll, so what is the probability of getting
the first 6 on the 4th roll of a die? Using our formula, we need to find small p and n. Small
p is the probability of getting a 6, which is 1 over 6, as there six possible outcomes on a die
roll. And n is equal to 4, because we're looking for the first success, getting a 6 on the 4th
roll of the die. So we have probability of 4 equals, 1 over 6, times 1, minus 1 over 6 raised
to the 4 minus 1. Calculating out, as I've done in the screen, we get 125 over 1296, which equals
0.096 or 9.6%. As another example let's say that your favorite
basketball player has a 70% free-throw percentage, what is the probability that the first free-throw
he makes occurs on the third or fourth attempt in tonight's game? So, in reading this question, we
see 3rd or 4th attempt, which means we're gonna have to calculate 2 probabilities, for
the 3rd attempt or third trial, and for the 4th attempt or fourth trial, and since this
question says 3rd or 4th, we will then add them together. Using our formula we need to find
small p and n, first for the 3rd attempt. We know that small p is a probability of success, that he
makes a free-throw and that is given, 70%, and n is equal to 3, because we're looking for the 1st
success on the 3rd free-throw attempt. Now we can plug and calculate. Probability of 3 equals,
0.70, times 1 minus 0.70 raised to the 3 minus 1. Do the math
and we get 0.063 or 6.3%. Now for the 4th attempt, we have
probability of 4 equals, 0.70, times 1 minus 0.70 raised to
the 4 minus 1. Calculate out and we get 0.019, or 1.9%. Next
we add these together, 0.019, plus 0.063, which equals
0.082, or 8.2%. So there's an 8.2% probability
that the player makes his 1st free-throw, on his 3rd or 4th attempt in tonight's
game. All right my friends, that be the basics on the geometric distribution. Hopefully this
video helped you out, I do have more videos right there for you, till next time, I am outta here.