Question

For a charity event, raffle tickets are printed such that they have a vowel (excluding y) as the first character and a prime number less than 10 as the second digit. (Note that 1 is not a prime number.) The probability that the winning ticket has the vowel contained in the word event and the digit is the number of letters in the word event is

Answer 1

Answer:

P=5%

Step-by-step explanation:

In order to solve this problem, we can use the following formula for the probability of two independent events:

P(A and B)=P(A)*P(B)

so we need to find the probability for each event to happen.

They are asking for the probability that the winning  ticket will have the letter e on it. There are 5 vowels in the alphabet (a, e, i, o and u). Therefore the probability for the first letter to be e is:

$P(A)=\frac{desired}{total}$

$P(A)=\rac{1}{5}$

We also know there are 4 prime numbers between 0 and 10 (2, 3, 5, 7) so the probability for the number to be a 5 is:

$P(B)=\frac{1}{4}$

so the probability of getting a ticket that starts with an e and ends with a 5 is:

P(A and B)=P(A)*P(B)

$P(A and B)=(\frac{1}{5})(\frac{1}{4})$

$P(A and B)=\frac{1}{20}$

so the probability is 5%