Question

In a word game, you choose a tile from a bag, replace it, and then choose another. If there are 28 vowels and 12 consonants, what is the probability you will choose a consonant and then a vowel?

Answer 1

Solutions :-

Given :
Number of vowels = 28
Number of consonants = 12
Total number of alphabets = (28 + 12) = 40


Find the probability of getting a vowel :-

We know that,
P(E) = F/T
P(getting a vowel) = Number of Vowels / Total number of Alphabets
= 28/40
= 7/10


Find the probability of getting a consonant :-

We know that,
P(E) = F/T
P(getting a consonant) = Number of consonants / Total number of Alphabets
= 12/40
= 3/10


Find the probability of getting a consonant and then a vowel (vice-versa) :-


P(getting a vowel) × P(getting a constant)
= (7/10) × (3/10)
= 21/100


Hence,
The probability of getting a consonant and then a vowel = 21/100

Answer 2

Answer:

21/100

Step By Step Explanation:

From The Condition, We can Conclude:

⇒ The numbers of vowels and consonants stay the same for the second tile, because the first tile was replaced.

Hence, The total number of tiles = 40.

Here, We take Probability , Consonant and Vowel as P , C & V respectively.

P(C,V) = P(C) × P(V) ________[i]

⇒ P(C,V) = 28 / 40 × 12 / 40

⇒ P(C,V) = 336 / 1600

⇒ P(C,V) = 21 / 100

Hence, The Required Probability is 21/100.