Determine the following probabilities
as percentages rounded to four decimal places.
For the first example,
Cameron is creating a seven-digit passcode
using the digits zero through nine.
If Cameron chooses all of the characters at random,
what is the probability that no digits are repeated?
Let's begin by working on the denominator,
which is equal to the total number of outcomes,
or in our case,
the total number of possible seven-digit passcodes
using the digits zero through nine,
if repetition is allowed.
So let's go ahead and model this.
We have one, two, three, four, five, six,
seven digit passcodes.
If repetition is allowed using accounting principle,
there are 10 ways to select the first digit,
10 ways to select the second digit,
10 ways to select a third digit and so on,
giving us a total of 10 to the seventh possible passcodes.
Again, when repetition is allowed.
So that's the denominator of the probability,
which is the total number of outcomes,
or total number of seven-digit passcodes using these digits.
Because we're looking for the probability
that no digits are repeated,
we now need to find how many passcodes are possible
if repetition is not allowed.
So if repetition is not allowed,
there are still 10 choices for the first digit,
but then since that digit cannot be used again,
there are only nine choices for the second digit,
eight choices for the third digit,
seven choices for the fourth digit,
six choices for the fifth digit,
five choices for the six digit,
and finally four choices for the seventh digit.
Again, this is if repetition is not allowed,
this product is equal to the favorable number of outcomes
or the numerator of the probability.
We could have also used a permutation here
and determined 10 permute seven,
but I think this approach makes more sense.
And now let's go to the calculator.
Let's click the fraction key,
and then we have 10 times nine times eight
times seven times six times five times four, Tab,
the denominator is 10,
and then we press the exponent key here,
enter the exponent of seven, and Enter.
To convert the decimal to a percent,
we multiply it by 100 and add a percent sign,
which gives us 6.048%.
We were told to round to four decimal places,
but here we have the exact percentage
in three decimal places.
For the second question,
three cards are chosen without replacement
from a deck of 52 cards
and placed in order from first to third.
What is the probability that all three cards are fives?
I think the easiest approach for this problem
will be to think of this as three dependent events.
And therefore, if I find the probability of each event,
we can simply multiply those probabilities together
to find the probability of three fives.
So we first find the probability
that the first card is a five.
Well, there are 52 cards, four of which are fives,
and therefore the probably the first card is a five
is four divided by 52 or 4/52
times, now we multiply by the probability
that the second card is a five,
given the first card is a five.
Well, now there's only 51 cards, three of which are fives.
And therefore the probability the second card is a five
given the first card is already a five
is three divided by 51, or 3/51.
And then we multiply by the probability
that third card is a five
given the first two cards are fives.
Well, now there's only 50 cards, two of which are fives,
and therefore the probability the third card is a five
given the first two cards are fives
is two divided by 50 or 2/50.
And now let's go to the calculator.
Four divided by 52, Tab, times three divided by a 51,
Tab, times two divided by 50, Enter.
Here we have the probability as a decimal,
to convert to a percent,
we multiply by 100 and add a percent sign.
So we have times 100.
And now we're told to round to four decimal places
because we have a nine in the fifth decimal place,
we round up to 0.0181%.
I hope you found this helpful.