Question 1061605: #1. 400 families were surveyed. It was found that 90% had a TV set & 60% had a computer. Every family had at least one of these items. One of these families is randomly selected & it is found that they have a computer. Find the probability that it also has a TV set.
#2. Two dice are rolled. What is the probability of getting;
a. Not a 3
b. The same # on both dice
c. A sum of 8
d. A sum of 3 or 4
e. A pair where the first # is less than the second #
f. A sum of 12
g. A sum of 20
h. 3,4 or 6,2
i. Two even #s
j. A sum less than 8
k. A sum greater than 8
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! P(TV or C)=P(TV)+p(C)-P(Both)
1=0.9+0.6-P(Both)
Therefore, P(Both)=0.5
Probability choose one with a computer is 0.6
If they have a TV, they fit into the category of having both.
Therefore, 0.6*P(TV)=0.5
P(TV)=0.5/0.5=0.83333 or 5/6
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For Dice, The probability of a 3 is 2/36, 36 ways to roll and 1/2 or 2/1. So it is 33/36=11/12.
The same number is 1/6, 11,22,33,44,55,66
A sum of 8 is 5/36. 2/6,3/5,4/4,5/3,6/2
A sum of 3 or 4 is 5/36:1/2,2/1,1/3,2/2,3/1
A pair where the first number is less than the second
1,2-1/6 (5); 2/3-2/6 (4), 3/4-3/6 (3), 4/5-4/6 (2), 5/6 (1). That is 15/36 or 5/12.
A sum of 12 is 1/36, 6/6
A sum of 20 is 0.
3,4 or 6,2 is 2/36 or 1/18, assuming order matters, otherwise it is 1/9.
Two even numbers are 2/2,4/4,6/6 or 3/36 or 1/12
A sum less than 8 is (1/36, 2/36,3/36,4/36,5/36,6/36)=21/36 or 7/12.
A sum greater than 8 is 4/36,3/36,2/36,1/36=10/36=5/18
If one makes a matrix of
1==2==3==4==5==6
2
3
4
5
6
and fills it in, all these probabilities can be easily seen. Seven is the most common sum, 1/6 times, with each number away from that sum being 1/36 less.
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