Question

The letters of the alphabet are written on 26 cards.Two cards are chosen at random without replacement. What is the probability that at least one of them is a consonant

Answer 1

Answer:

The probability of selecting at least one consonant when selecting 2 cards is 0.0954.

Step-by-step explanation:

There are 26 letters in the English alphabet series.

Of these 26 letters there are 5 vowels and 21 consonants.

The total number of ways to select 2 letters is:

$N={26\choose 2}=\frac{26!}{2!\times24!} =325$

The number of ways to select at least one consonant is:

n (At least 1 consonant) = N - n (Two vowels)

                                      $=325-{5\choose 2}\\=325-[\frac{5!}{3!\times2!}]\\=325-10\\=315$

Compute the probability of selecting at least one consonant as follows:

$P(At\ least\ 1\ consonant)=\frac{n(At\ least\ 1\ consonant)}{N} =\frac{315}{325}=0.0954$

Thus, the probability of selecting at least one consonant when selecting 2 cards is 0.0954.

Answer 2

Answer:

Step-by-step explanation:

There are 5 vowels in an alphabet and 21 consonants in an alphabet.

Case I: Both are the consonants:

Probability to choose 1 consonant = 21 /26

Now there are 25 cards remaining and 20 consonant

Probability to choose another consonant = 20/25

So, probability to choose 2 consonants

P' = $\frac{21}{26}\times\frac{20}{25}$

P' = 0.646

Case I: one is consonant and other is vowel:

Probability to choose 1 consonant = 21 /26

Now there are 25 cards remaining.

Probability to choose another card which is a vowel = 5/25

So, probability to choose 2 consonants

P'' = $\frac{21}{26}\times\frac{5}{25}$

P'' = 0.162

Total probability to choose at least one consonant

P = P' + P''

P =0.646 + 0.162

P = 0.808