Since the sum of all probabilities of all all elementary events will always be equal to 1. Furthermore, the probabilities of all mutually exclusive set of events that is part of the entire sample space will always be total of 1.
So in the problem, the answer is 1/8.
1/8 for red + 3/8 for green + 3/8 for yellow + 1/8 for blue = 8/8 or 1.
Answer:

Step-by-step explanation:
Given that from a well shuffled set of playing cards (52 in number) a card is drawn and without replacing it, next card is drawn.
A - the first card is 4
B - second card is ace
We have to find probability for

P(A) = no of 4s in the deck/total cards = 
After this first drawn if 4 is drawn, we have remaining 51 cards with 4 aces in it
P(B) = no of Aces in 51 cards/51 = 
Hence

(Here we see that A and B are independent once we adjust the number of cards. Also for both we multiply the probabilities)
Answer:
The probability that the next failure will not occur before 30 months have elapsed is 0.0454
Step-by-step explanation:
Using Poisson distribution where
t= number of units of time
x= number of occurrences in t units of time
λ= average number of occurrences per unit of time
P(x;λt) = e raise to power (-λt) multiplied by λtˣ divided by x!
here λt = 25
x= 30
P(x= 30) = 25³⁰e⁻²⁵/ 30!
P (x= 30) = 8.67 E41 * 1.3887 E-11/30! (where E= exponent)
P (x=30) = 1.204 E31/30!
Solving it with a statistical calculator would give
P (x=30) = 0.0454
The probability that the next failure will not occur before 30 months have elapsed is 0.0454
Answer:
A. $301
B. $721
Step-by-step explanation:
Let $x be the amount of money they raised.
Rowena tried to put the $1 bills into two equal piles and found one left over at the end, then

Polly tried to put the $1 bills into three equal piles and found one left over at the end, then

Frustrated, they tried 4, 5, and 6 equal piles and each time had $1 left over, then

Finally Rowena put all the bills evenly into 7 equal piles, and none were left over, then

This means
is divisible by 2, 3, 4, 5 and 6 without remainder, so

Hence,

The smallest amount of money they could have raised is $301, because
is divisible by 7.
Now, the number
should be divisible by 7 and must be greater than 500.
So,

When n = 9,
is not divisible by 7.
When n = 10,
is not divisible by 7.
When n = 11,
is not divisible by 7.
When n = 12,
is divisible by 7.
B. The least amount of money they could have raised is $721
Weight of an object=J
x=J/(379.2/150)
x=J/2.528
y=12.64/2.528
y=5
Hope this helps :)