Answer:
Random Sample
Step-by-step explanation:
Probability that you have chosen 2 red marbles is 3/11
<h2>How to solve?</h2>
Total number of marbles = 6 + 5 = 11
Number of red marbles = 6
Number of blue marbles = 5
Important Points:
1. Replacement is not allowed
2. Blue is followed by Red
Probability has been introduced in Maths to predict how likely events are to happen. The meaning of probability is basically the extent to which something is likely to happen.
Probability of Blue first = 5/11
Probability of Red second = 6/10 (one marble is already taken out)
Probability = 5/11 * 6/10 = 3/11
Learn more about Probability:
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Answer:
0.9855 or 98.55%.
Step-by-step explanation:
The probability of each individual match being flawed is p = 0.008. The probability that a matchbox will have one or fewer matches with a flaw is the same as the probability of a matchbox having exactly one or exactly zero matches with a flaw:

The probability that a matchbox will have one or fewer matches with a flaw is 0.9855 or 98.55%.
Problem 5
Apply the Law of Sines
s/sin(S) = r/sin(R)
s/sin(78) = 10/sin(48)
s = sin(78)*10/sin(48)
s = 13.162274
<h3>Answer: 13.162274 approximately</h3>
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Problem 6
Use the Law of Sines here as well.
x/sin(X) = y/sin(Y)
x/sin(53) = 6/sin(22)
x = sin(53)*6/sin(22)
x = 12.791588
<h3>Answer: 12.791588 approximately</h3>
Answer:
A) 
B) 
C) for n = 2
= 1
for n = 3
= 3
for n = 4
= 8
for n = 5
= 19
Step-by-step explanation:
A) A recurrence relation for the number of bit strings of length n that contain a pair of consecutive Os can be represented below
if a string (n ) ends with 00 for n-2 positions there are a pair of consecutive Os therefore there will be :
strings
therefore for n ≥ 2
The recurrence relation for the number of bit strings of length 'n' that contains consecutive Os
b ) The initial conditions
The initial conditions are : 
C) The number of bit strings of length seven containing two consecutive 0s
here we apply the re occurrence relation and the initial conditions
for n = 2
= 1
for n = 3
= 3
for n = 4
= 8
for n = 5
= 19