Calculating Probability of Choosing Vowels: Common Mistakes and Solutions

[songoku]

Homework Statement

One letter is randomly selected from each of the words HOORAY FOR MATHS. Find the probability of getting at least one vowel which is not the letter O

Relevant Equations

Probability

My attempt:

case 1: get one vowel (A) from word HOORAY = 1/6 x 2/3 x 4/5 = 4/45

case 2: get one vowel (A) from word MATHS = 3/6 x 2/3 x 1/5 = 1/15

case 3: get two vowels (2A) from word HOORAY and MATHS = 1/6 x 2/3 x 1/5 = 1/45

Total probability = 8/45

Answer key = 1/3

Where is my mistake?

Thanks

 

[PeroK]

Homework Statement: One letter is randomly selected from each of the words HOORAY FOR MATHS. Find the probability of getting at least one vowel which is not the letter O
Homework Equations: Probability

My attempt:

case 1: get one vowel (A) from word HOORAY = 1/6 x 2/3 x 4/5 = 4/45

case 2: get one vowel (A) from word MATHS = 3/6 x 2/3 x 1/5 = 1/15

case 3: get two vowels (2A) from word HOORAY and MATHS = 1/6 x 2/3 x 1/5 = 1/45

Total probability = 8/45

Answer key = 1/3

Where is my mistake?

Thanks

It's okay to get the letter O. If you get A-O-M, then that is a successful outcome.

But, it would have been quicker to caluculate the probability of not getting a non-O vowel.

 

[songoku]

It's okay to get the letter O. If you get A-O-M, then that is a successful outcome.

But, it would have been quicker to caluculate the probability of not getting a non-O vowel.

I see, so that is the meaning of the question. Thank you very much perok

 

[WWGD]

It seems you can use inclusion/exclusion here:$p(A \cup B \cup C)=p(A)+p(B)+p(C)-p(A \cap B)...$

 

[WWGD]

Homework Statement: One letter is randomly selected from each of the words HOORAY FOR MATHS. Find the probability of getting at least one vowel which is not the letter O
Homework Equations: Probability

My attempt:

case 1: get one vowel (A) from word HOORAY = 1/6 x 2/3 x 4/5 = 4/45

case 2: get one vowel (A) from word MATHS = 3/6 x 2/3 x 1/5 = 1/15

case 3: get two vowels (2A) from word HOORAY and MATHS = 1/6 x 2/3 x 1/5 = 1/45

Total probability = 8/45

Answer key = 1/3

Where is my mistake?

Thanks

For one, the probability of getting an A in the word 'For' is 0 (as well as in the HW if you don't change it ;)).

 

[HallsofIvy]

I may be interpreting this slightly differently. I would say that the "one vowel that is not O" MIGHT be the "A" from the first word, HOORAY. There are six letters, one of which is "A" so the probability of that is 1/6. There is no "vowel that is not O" in the second word "FOR" but there is one "A" in the word MATHS. The probability of getting that is 1/5 so the probability of getting "at least one vowel that is not O" is 1/6+ 1/5= 11/30.

 

[PeroK]

I may be interpreting this slightly differently. I would say that the "one vowel that is not O" MIGHT be the "A" from the first word, HOORAY. There are six letters, one of which is "A" so the probability of that is 1/6. There is no "vowel that is not O" in the second word "FOR" but there is one "A" in the word MATHS. The probability of getting that is 1/5 so the probability of getting "at least one vowel that is not O" is 1/6+ 1/5= 11/30.

That double counts the case when you get an "A" from both the first and third words. This happens with probability 1/30. If you subtract that you get 10/30, as expected.