How many different committees of 4 people can be chosen from a group of 21?
1 answer:
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the complete question in the attached figure
1) Find the probability that all nine victims would select the same person. You should enter your answer as a decimal, not as a percent. Do not round, as your answer should be a terminating decimal.
P=(1/5)*(1/5)*(1/5)*(1/5)*(1/5)*(1/5)*(1/5)*(1/5)*(1/5)=0.000000512
2) what is the probability that the first victim picks SOMEONE? ANYONE?
P=(1/5)=0.20 ---------------- > 20%
3) Now let’s look at victim #2. What is the probability that they pick the same person as victim #1?
P=(1/5)*(1/5)=0.04 --------------> 4%
4) Now let’s look at victim #3. What is the probability that they pick the same person as victim #1?
P=(1/5)*(1/5)*(1/5)=0.008 --------------> 0.8%
5) Now let's look at victim #9. What is the probability that they pick the same person as victim #1?
P=(1/5)*(1/5)*(1/5)*(1/5)*(1/5)*(1/5)*(1/5)*(1/5)*(1/5)=0.000000512
6) Once you find all those probabilities, what rule would apply?
the probability is given by (1/5)^n
n=number of victims that pick the same person---------> in this problem = 9
P=(1/5)^9= 0.000000512
Answer:
4th one
Step-by-step explanation:
a^-2 goes down and become a^2 and
b^-1 goes up and become b
that why answer is 4th...mark me brainliest please