SOLUTION: Three cards labelled 2, 3 and 4 are placed in a hat and a card is drawn. This is repeated 3 times without replacement. a) How many 3 digit numbers can be formed b) What is the p

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Question 1129131: Three cards labelled 2, 3 and 4 are placed in a hat and a card is drawn. This is repeated 3 times without replacement.
a) How many 3 digit numbers can be formed
b) What is the probability that the 3 digit number is odd? divisible by 3?
Found 2 solutions by htmentor, Edwin McCravy:
Answer by htmentor(1343)   (Show Source): You can put this solution on YOUR website!
Since the cards are drawn without replacement, there are 3 possibilities for
the 1st number, 2 possibilities for the 2nd number and 1 possibility for the
remaining number.
Thus, there are 3x2x1 = 6 combinations
Since the digits 2, 3 and 4 add up to 9, all the numbers are divisible by 9
and therefore also divisible by 3.
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
Three cards labelled 2, 3 and 4 are placed in a hat and a card is drawn. This is repeated 3 times without replacement.
a) How many 3 digit numbers can be formed?
These six: 234, 243, 324, 342, 423 and 432

b) What is the probability that the 3 digit number is odd? 
Only 243 and 423 are odd. That's 2 ways out of 6, or 2/6 which reduces to 1/3.

divisible by 3?
234÷3 = 78, so 234 is divisible by 3.
243÷3 = 81, so 243 is divisible by 3.
324÷3 = 108, so 324 is divisible by 3.
342÷3 = 114, so 342 is divisible by 3.
423÷3 = 141, so 423 is divisible by 3.
432÷3 = 144, so 432 is divisible by 3.

That's 6 ways out of 6, or 6/6 which reduces to 1. So it is certain that any
one of the 6 three-digit numbers you pick, that it will be divisible by 3. A
probability of an event being 1 means that the event is certain.

Edwin
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