There are two possible sums that add up to a number evenly divisible by 2 and 3, namely 6 and 12.
There are 5 different ways to roll a sum of 6, so the probability of that happening is 5/36.
There is 1 way to roll a sum of 12, so the probability of that happening is 1/36.
The probability of either happening is the sum of the two separate probabilities.
p=5/36+1/36=6/36=1/6 ~= .1667 = 16.67%
1/4
2/9!
The answer depends on what the experiment consists of. Since that is not specified, there can be no sensible answer.
The probability of flipping a Head is the same as that for a Tail and is 1/2 or 50%. The probability of rolling a particular number on a die is 1/6 since there are 6 numbers. Combining these two probabilities (by multiplication) we have, as the combined probability 1/2 x 1/6 = 1/12 = 0.0833333333333333(the 3 recurs) which as a percentage is 8.33333333333%
The sample space consists of the letters of the word "PROBABILITY" = {P,R,O,B,A,I,L,T,Y}
1/4
2/9!
The answer depends on what the experiment consists of. Since that is not specified, there can be no sensible answer.
If the family consists of a pair of parents and their offspring, then the probability is very close to 0.
The probability of flipping a Head is the same as that for a Tail and is 1/2 or 50%. The probability of rolling a particular number on a die is 1/6 since there are 6 numbers. Combining these two probabilities (by multiplication) we have, as the combined probability 1/2 x 1/6 = 1/12 = 0.0833333333333333(the 3 recurs) which as a percentage is 8.33333333333%
A binomial experiment is a experiment that consists of repeated trails, with two possible outcomes. An example of this would be a coin toss.Ê
False. The question consists of two parts: - a number is divisible by 6 if it is divisible by 3? False. It must also be divisible by 2. - a number is divisible by 6 only if it is divisible by 3? This is true but the false part makes the whole statement false.
The sample space consists of the letters of the word "PROBABILITY" = {P,R,O,B,A,I,L,T,Y}
9873
The probability of a woman being chosen in this situation is 7/11.
9867
The probability histogram consists of eight bars, labeled '1' through '8', each bar having a value (height) of 0.125 .