Ask your own question, for FREE!
Mathematics 38 Online
OpenStudy (anonymous):

You have 26 cards, each labeled with a letter of the alphabet. What is the probability of drawing a card that contains one of the letters in the word 'MATH'? A) 1 676 B) 2 13 C) 2 26 D) 3 26

OpenStudy (anonymous):

1/676, 2/13, 2/26, and 3/26? those are the options?

OpenStudy (anonymous):

yes, sorry.

OpenStudy (anonymous):

in 'MATH' there are 4 unique letters. there are also 26 unique letters in the alphabet. Therefore, when you are choosing just one card, you have a 1/26 chance of selecting a given letter....so this question is only about choosing 1 card, but it's acceptable for that card to be M or A or T or H. so each letter has 1/26 chance of being selecting. because it's OK to get any of the letters in MATH, you have a better chance of obtaining a letter that you want. so you have the probability of obtaining a "M" plus that of obtaining an "A" plus...

OpenStudy (anonymous):

so, of the answer choices, which will you choose?

OpenStudy (anonymous):

:s

OpenStudy (anonymous):

all right, let's start with this idea. If I wanted to pull an "A" card, what is the probability of obtaining the "A"? there is only 1 A in all of the 26 cards

OpenStudy (anonymous):

1/26

OpenStudy (anonymous):

right

OpenStudy (anonymous):

ok, so for any 1 letter, the probability of obtaining it is 1/26.

OpenStudy (anonymous):

mhm

OpenStudy (anonymous):

ok, for the question, you can select 1 card, but it's OK if the letter is equal to any of the 4 in "MATH"

OpenStudy (anonymous):

so 4/26?

OpenStudy (anonymous):

imagine that I want to select 1 card, and I will accept the card if it is any letter EXCEPT "A", therefore, there is 1/26 probability of selecting "A", and a (1-1/26) = 25/26 probability of selecting any other letter....

OpenStudy (anonymous):

nice, how did you get that, 4/26?

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!