Answer:
27/200
Step-by-step explanation:
Knowing that:
A bag of tokens contains 9 red, 6 green, and 5 blue tokens. What is the probability that a randomly selected token is red and green?
Solve:
and = multiply
Total = 9 red + 6 green + 5 blue = 9 + 6 + 5 = 20
Total = 20 token
Thus,
There are 9 red and 6 green..
Hence,
9 out of 20 is red
6 out of 20 is green
Equation
9/20 × 6/20
Multiply Across:
9 × 6 = 54
20 × 20 = 400
54/400
Simplify
27/200
Therefore, the probability that a randomly selected token is red and green is 27/200.
<u><em>~Lenvy~</em></u>
First we need to make both fractions equal to each other, and that means having to have the denominators the same.
What both 3 and 12 easily go into is 12.
So we need to make both fractions have a denominator of 12, so we multiply.
2/3 * 4/4 = 8 / 12
1/4 * 3/3 = 3 / 12
Now, we know that she has 8 / 12 pounds, and she uses 3 / 12 of it, so now we subtract to find how much she has left.
8 - 3 = 5
Out of the 12 pounds of roast beef, she now has 5 / 12 left.
Answer:
a) P(6) = 0.0097
b) P(More than 3) = 0.1611
Step-by-step explanation:
For each question, there are only two possible outcomes. Either it is guessed correctly, or it is not. Questions are independent of each other. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.
A student takes a multiple-choice test that has 11 questions.
This means that 
Each question has five choices.
This means that 
(a) Find P (6)
This is P(X = 6).


P(6) = 0.0097
(b) Find P (More than 3).
Either P is 3 or less, or it is more than three. The sum of the probabilities of these outcomes is 1. So

We want P(X > 3). So

In which







Then

P(More than 3) = 0.1611
B? I’m not quite sure but I think it’s that.