Write the set of points from -6 to 0 but excluding -4 and 0 as a union of intervals
First we take the interval -6 to 0. In that -4 and 0 are excluded.
So we split the interval -6 to 0.
Start with -6 and go up to -4. -4 is excluded so we break at -4. Also we use parenthesis for -4.
Interval becomes [-6,-4) . It says -6 included but -4 excluded.
Next interval starts at -4 and ends at 0. -4 and 0 are excluded so we use parenthesis not square brackets
(-4,0)
Now we take union of both intervals
[-6,-4) U (-4,0) --- Interval from -6 to 0 but excluding -4 and 0
The simulation of the medicine and the bowler hat are illustrations of probability
- The probability that the medicine is effective on at least two is 0.767
- The probability that the medicine is effective on none is 0
- The probability that the bowler hits a headpin 4 out of 5 times is 0.3281
<h3>The probability that the medicine is effective on at least two</h3>
From the question,
- Numbers 1 to 7 represents the medicine being effective
- 0, 8 and 9 represents the medicine not being effective
From the simulation, 23 of the 30 randomly generated numbers show that the medicine is effective on at least two
So, the probability is:
p = 23/30
p = 0.767
Hence, the probability that the medicine is effective on at least two is 0.767
<h3>The probability that the medicine is effective on none</h3>
From the simulation, 0 of the 30 randomly generated numbers show that the medicine is effective on none
So, the probability is:
p = 0/30
p = 0
Hence, the probability that the medicine is effective on none is 0
<h3>The probability a bowler hits a headpin</h3>
The probability of hitting a headpin is:
p = 90%
The probability a bowler hits a headpin 4 out of 5 times is:
P(x) = nCx * p^x * (1 - p)^(n - x)
So, we have:
P(4) = 5C4 * (90%)^4 * (1 - 90%)^1
P(4) = 0.3281
Hence, the probability that the bowler hits a headpin 4 out of 5 times is 0.3281
Read more about probabilities at:
brainly.com/question/25870256
Answer: a) 
b) 
<u>Step-by-step explanation:</u>
a) In order to get an even number, you have the 3 different scenarios:
1) Even, Even, Even, Even 
2) Even, Even, Odd, Odd
3) Odd, Odd, Odd, Odd
<em>Order doesn't matter</em>
Add them up to get your answer: 
b) If one die is a 2 and another is a 3 and the other two dice can be any number, then you have 1 possibility for a 2, 1 possibility for a 3, and 6 possibilities for each of the other two dice.
