Find the probability of picking 1 consonant and 4 vowels when 5 letters are picked (without replacement) from a set of alphabet tiles.

Ok, we have a total of 26 letters, and we want to select 5 of them.

Of the 26 letters, 21 are consonants and 5 are vowels.

Suppose that we want to have the consonant in the first selection, so the probability of picking a consonant is equal to the quotient between the number of consonants and the total number of letters.

p = 21/26

now a letter has been selected, so in the set, we have 25 letters left.

In the next 4 selections, we must select vowels.

In the second selection the probability is:

p = 5/25

in the third, the prob is:

p = 4/24 (we already selected one vowel before, so now we only have 4 vowels)

The fourth selection:

p = 3/23

and the last selection:

p = 2/22

The total probability is equal to the product of all the individual proabilities, so we have:

P = (2/22)*(3/23)*(4/24)*(5/25)*(21/26)

Now, remember that we said that the consonant must be in the first place, but it can be in any of the five places, so if we add the permutations of the consonant letter we have:

P = 5*(2/22)*(3/23)*(4/24)*(5/25)*(21/26) =  0.0018

last option

step-by-step explanation:

given that δdef is right and isoceles,

assuming that side df is the hypotenuse means that de an ef are the other 2 sides that intersect at 90° and are of equal length.

because they are of equal length, we can say that they are congruent

because they intersect at 90°, their slopes must be opposite reciprocals.

only the last option satisfies these conditions.