Find the indicated probability. Find the probability that 3 randomly selected people all have the same birthday. Ignore leap years. Round to eight decimal places

P = 0.00000751

the probability that 3 randomly selected people all have the same birthday is 0.00000751

Step-by-step explanation:

the probability that 3 randomly selected people all have the same birthday P;

P = P1 × P2 × P3 1

For the first person, since there is no specified date, he can have any of the 365 days of the year as his birthday. the probability P1 is;

P1 = 365/365 = 1

For the second person, he must have the same birthday as the first. The probability P2 is;

P2 = 1/365

The third person must also have the same birthday as the other two. The probability P3 is;

P3 = 1/365

Substituting the values into equation 1;

P = 1 × 1/365 × 1/365

P = 0.0000075060987051

To eight decimal places;

P = 0.00000751

the probability that 3 randomly selected people all have the same birthday is 0.00000751

c. 16 units

step-by-step explanation:

mark brainliest and have a great day!

d

step-by-step explanation:

hello and welcome!

so, when we see the words "proportional" along with a table, we need to know that we're looking numbers that divide by the same thing.

looking at this table, we simply have to do a trial and error, and divide to see which numbers have the same quotient. if you look at choice a & c, they both have equal quotients for three of the numbers, with one exception, so those can't be the answer. looking at b, there is no similarity, so that can't be it. then, we see d; if you divide all the numbers in the table, you'll see that they all divide by two, which is a proportional relationship, and our answer!

also, always remember that the x will always be the number that goes into y, and somewhere behind this table, is a formula. for this specific table, the formula would be:

since our x-value is being divided by two to get the y-value. and if they were to ask for more numbers in the series, this is the formula you would use to find them.