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Sagot :
Answer:
[tex]P(X = 0) = 0.03125[/tex]
[tex]P(X = 1) = 0.15625[/tex]
[tex]P(X = 2) = 0.3125[/tex]
[tex]P(X = 3) = 0.3125[/tex]
[tex]P(X = 4) = 0.15625[/tex]
[tex]P(X = 5) = 0.03125[/tex]
Step-by-step explanation:
For each toss, there are only two possible outcomes. Either it is tails, or it is not. The probability of a toss resulting in tails is independent of any other toss, which means that the binomial probability distribution is used 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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Fair coin:
Equally as likely to be heads or tails, so [tex]p = 0.5[/tex]
5 tosses:
This means that [tex]n = 5[/tex]
Probability distribution:
Probability of each outcome, so:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{5,0}.(0.5)^{0}.(0.5)^{5} = 0.03125[/tex]
[tex]P(X = 1) = C_{5,1}.(0.5)^{1}.(0.5)^{4} = 0.15625[/tex]
[tex]P(X = 2) = C_{5,2}.(0.5)^{2}.(0.5)^{3} = 0.3125[/tex]
[tex]P(X = 3) = C_{5,3}.(0.5)^{3}.(0.5)^{2} = 0.3125[/tex]
[tex]P(X = 4) = C_{5,4}.(0.5)^{4}.(0.5)^{1} = 0.15625[/tex]
[tex]P(X = 5) = C_{5,5}.(0.5)^{5}.(0.5)^{0} = 0.03125[/tex]
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