Mathematics!

Probability Level 4

I choose 2 letters from the word MATHEMATICS. The first letter was a vowel. The probability that the other letter is A can be represented as a b \frac {a}{b} , where a a and b b are co-prime positive integers. Find a + b a + b .


The answer is 23.

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Sharky Kesa by Sharky Kesa , 19, United Kingdom

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3 solutions

There are four particular cases that satisfy the condition: two AAs, EA, and IA. Probability of each case:

P ( A A ) = 1 4 ∗ 1 10 ∗ 2 = 1 20 P(AA)=\frac{1}{4}*\frac {1}{10}*2=\frac {1} {20}

P ( E A ) = 1 4 ∗ 2 10 = 1 20 P(EA)=\frac{1}{4}*\frac{2}{10}=\frac {1} {20}

P ( I A ) = 1 4 ∗ 2 10 = 1 20 P(IA)=\frac{1} {4}*\frac {2} {10}=\frac {1} {20}

So the total probability is P ( T o t a l ) = 1 20 + 1 20 + 1 20 = 3 20 P(Total)=\frac{1}{20}+\frac{1}{20}+\frac{1}{20}=\frac{3}{20}

Hence, a + b = 23 a+b=23

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Chamelean Brown
Oct 12, 2014

If letter #1 is not an A, then the probability is 2 10 \frac{2}{10} .

If letter #1 IS an A, then the probability is 1 10 \frac{1}{10} .

The total probability is the average of the two eventualities, as the probability of letter #1 being an A is 1 2 \frac{1}{2} .

1 2 ( 2 10 + 1 10 ) = 3 20 = a b \frac{1}{2}(\frac{2}{10}+\frac{1}{10}) = \frac{3}{20} = \frac{a}{b} a + b = 23 a+b = \boxed{23}

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The probability that an A is selected as the first letter is 2/11. If an A is selected first, then the probability that the second is an A as well is 1/10. The probability of AA is thus 2/11 * 1/10 = 2/110.

Similarly, the probability that an (E or I) is selected first is 2/11 and the probability that the second is an A is 2/10. Thus the probability of (EA or IA) is 2/11*2/10 = 4/110. The total probability is 4/110 + 2/110 = 3/55. The answer is 58.

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Please may you explain why the condition that 'the first letter was a vowel' does not affect the probability. I thought that having got 3/55, we then had to divide through by the chance of the given condition; that is 4/11, that the first letter is a vowel. Therefore I thought that the answer was 3 55 ÷ 4 11 = 3 20 \frac{3}{55} \div \frac{4}{11} = \boxed{\frac{3}{20}}

Michael Ng - 6 years, 8 months ago

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This is exactly what my argument is. The answer must be 23. Nice solution @Michael Ng

Kunal Jadhav - 6 years, 8 months ago

The problem tells you that the first letter is a vowel, so we are only finding the conditional probability that the second letter is an A given that the first is a vowel, not the probability that the second letter is an A in general.

Narahari Bharadwaj - 6 years, 8 months ago

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This comment is correct. However, it contradicts your solution.

Given that the first letter is a vowel, we know that the probability that the first letter is an A is 2 4 \frac{2}{4} and not 2 11 \frac{2}{11} .

What you did, was calculate the probability that the first letter is a vowel, and that the second letter is an A.

Calvin Lin Staff - 6 years, 8 months ago

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You are absolutely right. My apologies.

Narahari Bharadwaj - 6 years, 8 months ago

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@Narahari Bharadwaj – Thanks. I have updated the answer to 23.

Calvin Lin Staff - 6 years, 8 months ago

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@Calvin Lin – Thanks Calvin! :)

Kunal Jadhav - 6 years, 8 months ago

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