Question

Suppose that the letter cards for the word MATHEMATICS were put face down and mixed up and a card is picked up at random. What is the probability of picking up a vowel?

Answer 1

Hint: We will first calculate the number of total letters in the word ‘MATHEMATICS’ and then calculate the number of vowels in it, then put these values in the formula of probability of an event and get the required answer.

Complete step-by-step answer:
We are given the word “MATHEMATICS”.
We have 2 M’s, 2 A’s, 2 T’s, 1 H, 1 E, 1 I, 1 C and 1 S in it.
So, in total we have (2 + 2 + 2 + 1 + 1 +1 + 1 + 1) letters in it which is equivalent to 11 letters.
So, if the event is picking a card of letters among the letters contained in the word “MATHEMATICS”, then the number of points in our sample space is 11.
So, n(S) = 11, where n(S) means cardinality of S that is equivalent to saying number of sample points.
Now, let us see the number of vowels in the word “MATHEMATICS”.
We have 2 A’s, 1 E and 1 I as vowels in it.
So, the total number of vowels in the word “MATHEMATICS” is (2 + 1 + 1) letters that is equal to 4 letters.
Now, let us say ‘E’ is the event that when we pick a card randomly, we get a vowel.
Then, n(E) = 4, where n(E) represents the number of points in ‘E’.
The formula of probability of an event ‘E’ is given by:- \[P(E) = \dfrac{{n(E)}}{{n(S)}}\].
So, here putting in the values, we just calculated, we get:-
\[P(E) = \dfrac{4}{{11}}\].
Hence, the required probability is $\dfrac{4}{{11}}$

Note: The students must remember that here we have multiple of some letters like M and A but we calculate them differently, because we have put in all the letters of MATHEMATICS in all. So, we need to consider them differently only. But always remember that if the question would have been that you have the set of letters in MATHEMATICS, then you need to consider every letter once only because a set contains one element once only, duplicates cannot exist in the sets.