+0

# Probability Challenge.

0
721
3

A die is rolled three times what is the probability of getting three 5's? P(three 5's).

Three letters are selected at random from the word "distance" find the probability of three vowels P(3 vowels).

Two cards are drawn from a standard deck of 52 cards without replacement find the probability that both cards are aces.

From a group of 7 men and 5 women, a 4-person committee is chosen. what is the probability that all 4 are men or all 4 are women.

Each of the numbers 1 to 20 is written on a card and placed in a bag. If one card is drawn at random, what is the probability that it is a multiple of 3 or a multiple of 5.

How many different possible arrangements of the letters in the word "Indiana" are possible.

Jun 12, 2014

#3
+890
+5

Anonymous' method is correct as well, (it's probability method or counting method again), but he/she should learn to question results given by a calculator, 1/2016 is absurd.

Anonymous also gets the third question wrong. The question states 'without replacement'. That means that the second probability should be 3/51, not 3/52.

Jun 13, 2014

#1
+5

A dice is rolled three times what is the probability of getting three 5's?  P(three 5's) 1/216 (1/6*1/6*1/6)

Three letters are selected at random from the word "distance" find the probability of three vowels P(3 vowels)
3/8 x 2/7 x 1/6 = 1/2016

•
Two cards are drawn from a standard deck of 52 cards without replacement find the probability that both cards are aces
4/52 x 3/52 = 3/676

From a group of 7 men and 5 women, a 4-person committee is chosen. what is the probability that all 4 are men or all 4 are women.
`p(all men) = 7/12*6/11*5/10*4/9 = 7/99`
` P(all women) = 5/12*4/11*3/10*2/9 = 1/99`
Each of the numbers 1 to 20 is written on a card and placed in a bag. if one card is drawn at random, what is the probability that it is a multiple of 3 or a multiple of 5.
6/20+4/20-1/20 = 9/20

how many different possible arrangements of the letters in the word Indiana are possible 630
Jun 12, 2014
#2
+96302
+5

Three letters are selected at random from the word "distance" find the probability of three vowels P(3 vowels).

..........................................................................................................................................................

I see this one a little differently......we are counting all the sets that are possible when choosing any three things from eight =  C(8,3) = 56

Only one of these sets contain all three vowels in the word "distance."

So we have......... 1/56

----------------------------------------------------------------------------------------------------------------------------

Can someone look at my answer and tell me if I'm correct, or not??

Jun 12, 2014
#3
+890
+5