Respuesta :
Given: Five letter codes are to be made with the letter A, F, E, R and M without repitition.
To Find: Probability that a randomly chosen code starts with M and ends with E.
Solution: So, required probability is [tex]\frac{1}{20}[/tex].
Calculation/Explanation:
Let us calculate the required probability step-wise.
To pick the first letter as M, there are 5 possible outcomes (A, F, E, R or M) and picking M is only one possibility out 5 outcomes.
So, probability of picking an M first is equal to [tex]\frac{1}{5}[/tex] ...(1)
After this, in the second, third and fourth places, we must not pick an E. In the second place, there are 4 possible outcomes (A, F, E or R) out of which not picking an E means there are 3 possibilities (A, F or R) out of the 4 outcomes. So, for the second letter the probability of not picking an E is [tex]\frac{3}{4}[/tex] ...(2)
For the third letter, there are now 3 possible outcomes (since out 5 letters, 2 have already been chosen). One of these outcomes is an E and two will be non-E. This means, probability of not drawing an E is [tex]\frac{2}{3}[/tex] ...(3)
For the fourth letter, there are now 2 possible outcomes out of which one is an E and one is not an E. This means, the probability of not picking an E is [tex]\frac{1}{2}[/tex] ...(4)
For the final letter, only one possible outcome remains which is E. And therefore the probability that we draw an E would be 1 ...(5)
To get the total probability of starting the code with M and ending with E, we multiply the individual probabilities i.e., (1), (2), (3), (4) and (5).
So, required probability is [tex]\frac{1}{20}[/tex].
The probability that a randomly chosen code starts with M and ends with an E is 0.05
The letters are given as:
- A, F, E, R and M
The number of characters is 5.
From the question, we have:
- The first character can only be M (1)
- The last character can only be E (1)
- The second character can be any of A, F and R (3)
- The third character can be any of the remaining 2
- The fourth character can only be 1
The number of possible password from the above highlights is:
[tex]Password = 1 \times 1 \times 3\times 2 \times 1[/tex]
[tex]Password = 6[/tex]
The total number of password, without restriction is:
[tex]Total = 5![/tex]
This gives
[tex]Total = 120[/tex]
The required probability is then calculated as:
[tex]P = \frac{6}{120}[/tex]
Express as decimal
[tex]P = 0.05[/tex]
Hence, the probability that a randomly chosen code starts with M and ends with an E is 0.05
Read more about probability at:
https://brainly.com/question/251701
