7. How many different signals can be made by displaying six flags, all at a time, on a flagpole, if three are white, one blue, one red, and one yellow?
-- Here, I thought you could to 6! to equal 720 possible combinations, but my teacher marked this problem wrong.
14. Two dice are rolled. What is the probability that either the sum of the numbers showing is 7 or both numbers are even?
-- I do not understand how to do this problem at all :(
------------------------------------------------------------------------------
Can someone help me with these problems? Any help is greatly appreciated :) thanks!!
7) 6!/3!= 120
There are 6 ways to get a 7
there are 9 ways to have both even
these are mutually exclusive events, that is, you can't have a sum of 7 AND both be even
so Prob(a sum of seven OR both are even)
= 16/36 = 4/9
or by formula
Prob(sum of seven OR both even)
= Prob(sum of seven) + prob(both even) - prob(sum of seven AND both even)
= 6/36 + 9/36 - 0
= 4/9
Ooooohhhhh I totally get it now!! Thanks anonymous and Reiny :)
Certainly! I'll be happy to help you with these problems.
7. To find the number of different signals that can be made by displaying six flags on a flagpole, we need to consider the arrangement of the colored flags. Since three flags are white, one is blue, one is red, and one is yellow, we can use the concept of permutations.
The total number of flags is 6, and out of these, we have three white flags. Therefore, the number of ways we can arrange the white flags among themselves is given by 3! (3 factorial), which is equal to 3 × 2 × 1 = 6.
After arranging the white flags, we have three remaining flags (one blue, one red, and one yellow) to be placed. The number of ways we can arrange these three flags among themselves is 3! = 3 × 2 × 1 = 6.
So, the total number of different signals that can be made is equal to the product of the two arrangements we calculated: 6 × 6 = 36.
Therefore, the correct answer is 36, not 720.
14. To find the probability that either the sum of the numbers showing on two rolled dice is 7 or both numbers are even, we need to consider the favorable outcomes and the total possible outcomes.
Favorable outcomes:
- The sum of the numbers showing on the dice is 7: This can happen in 6 different ways (1+6, 2+5, 3+4, 4+3, 5+2, 6+1).
- Both numbers are even: This can happen in 9 different ways (2+2, 2+4, 2+6, 4+2, 4+4, 4+6, 6+2, 6+4, 6+6).
Total possible outcomes: When two dice are rolled, each dice has 6 possible outcomes. So the total possible outcomes are 6 × 6 = 36.
Now, we need to calculate the probability by dividing the number of favorable outcomes by the total possible outcomes.
Probability = (Number of Favorable Outcomes) / (Total Possible Outcomes)
= (6 + 9) / 36
= 15 / 36
= 5 / 12
Therefore, the probability that either the sum of the numbers showing is 7 or both numbers are even is 5/12.
I hope this helps you understand how to approach these problems! If you have any more questions, feel free to ask.