At at least one die come up a 3?We can do this two ways:) The straightforward way is as follows. To get at least one 3, would be consistent with the following three mutually exclusive outcomes:the 1st die is a 3 and the 2nd is not: prob = (1/6)x(5/6)=5/36the 1st die is not a 3 and the 2nd is: prob = (5/6)x((1/6)=5/36both the 1st and 2nd come up 3: prob = (1/6)x(1/6)=1/36sum of the above three cases is prob for at least one 3, p = 11/36ii) A faster way is as follows: prob at least one 3 = 1 - (prob no 3's)The probability to get no 3's is (5/6)x(5/6) = 25/36.So the probability to get at least one 3 is, p = 1 - (25/36) = 11/362) What is the probability that a card drawn at random from an ordinary 52 deck of playing cards is a queen or a heart?There are 4 queens and 13 hearts, so the probability to draw a queen is4/52 and the probability to draw a heart is 13/52. But the probability to draw a queen or a heart is NOT the sum 4/52 + 13/52. This is because drawing a queen and drawing a heart are not mutually exclusive outcomes - the queen of hearts can meet both criteria! The number of cards which meet the criteria of being either a queen or a heart is only 16 - the 4 queens and the 12 remaining hearts which are not a queen. So the probability to draw a queen or a heart is 16/52 = 4/13.3) Five coins are tossed. What is the probability that the number of heads exceeds the number of tails?We can divide
Answer:
$40
Step-by-step explanation:
6 = .15x
6 / .15 = 40
Answer:
d
Step-by-step explanation:
hchfjtjjgjhkgiyggb bbhhu
There are two ways to work this out: normal variables or using "imaginary" numbers.
Normal variables:
![(7+2i)(3-i)\\(7*3)+[7*(-i)]+(3*2i)+[2i*(-i)]\\21-7i+6i-2i^{2}\\\\21-i-2i^{2}](https://tex.z-dn.net/?f=%20%287%2B2i%29%283-i%29%5C%5C%287%2A3%29%2B%5B7%2A%28-i%29%5D%2B%283%2A2i%29%2B%5B2i%2A%28-i%29%5D%5C%5C21-7i%2B6i-2i%5E%7B2%7D%5C%5C%5C%5C21-i-2i%5E%7B2%7D)
Imaginary numbers:
Using the result from earlier:
Now since
, then the expression becomes:
