The letters of the alphabet are written on 26 cards. Two cards are chosen at random. Find the probability that at least one of them is a vowel ?

Engineering

Mathematics

Conditional Probability

Question

$\frac{23}{65}$

$\frac{13}{24}$

$\frac{5}{65}$

$\frac{1}{23}$

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

The complement of A' at least one of them is vowels is neither of them is a vowel.

There are 5 vowels and 21 are consonants P (The first card chosen is not a vowel)

= \frac{21}{26}

P (The second card chosen is also not a vowel)

= \frac{20}{25}

P (Neither of them is a vowel)

= \frac{21}{26} \times \frac{20}{25} = \frac{42}{65}

P (At least one of them is a vowel)

= 1 - \frac{42}{65} = \frac{23}{65}

Please subscribe our Youtube channel to unlock this solution.