Respuesta :
Answer: a.) 0.999972
b.) 0.999719
Step-by-step explanation:
For a visiting student to contract the virus, the probability of success, p, is given as 60% = 0.6
This means probability of failure, q, is 40% = 0.4.
Number of sample n is given as 15.
To solve this question, we use the probability distribution Formula for selection. By this, the probability is given as
P(x=r) = nCr * p^r * q^n-r.
a. WhereP( x > 1) = 1 - P(x </= 1) i.e when x= 0 and 1
When x=0,
P(x=0) = 15C0 * 0.6^0 * 0.4 ^ 15.
P(x=0) = 1 * 1 * 0.000001 = 0.000001.
When x=1
P(x=1) = 15C1 * 0.6^1 * 0.4^14
P(x=1) = 15 * 0.6 * 0.000003
P(x=1) = 0.000027
Hence, probability of x</= 1 is given as
P(x=0) + P(x=1)= 0.000001 + 0.000027 = 0.000028
Hence, probability of x>1 = P(x>1) = 1-0.000028
P(x>1) = 0.999972.
Probability that more than one student contracts the virus = 0.999972.
b.) probability that more that 2student contracts the virus = P(x>2)
Like the previous solution, we determine for probability of when x less than or equal to 2 i.e when p(x</= 2), then subtract our answer from 1.
P(x>2) = 1 - P(x</= 2)
Since we know P(x=0) and P(x=1) from our previous solution, we just find for the P(x=2)
P(x=2) = 15C2 * 0.6^2 * 0.4^13
P(x=2) = 105 * 0.36 * 0.0000067
P(x=2) = 0.000253.
P(x</=2) = p(x=0) + p(x=1) + p(x=2)
P(x</=2) = 0.000001 + 0.000027 + 0.000253 = 0.000281
Hence probability that more than 2 students contract the virus, that is:
P(x>2) = 1 - 0.000281
P(x>2) = 0.999719
A) Probability that more than one student contracts the virus is;
P( x > 1) = 0.999975
B) Probability that more than 2 students contract the virus is:
P(x > 2) = 0.999721
We are told that the of contracting a stomach virus while visiting Mexico is 60%. Thus;
probability of success; p = 60% = 0.6
probability of failure; q = 1 - p
q = 1 - 0.6
q = 0.4
Sample size is; n = 15
This is the binomial probability distribution which has the general formula;
P(X = x) = ⁿCₓ × pˣ × q⁽ⁿ ⁻ ˣ⁾
A) Probability that more than 1 student contacts the virus;
P( x > 1) = 1 - (P(x = 1) + P(x = 0))
P(x = 0) = ¹⁵C₀ 0.6⁰ 0.4⁽¹⁵ ⁻ ⁰⁾
P(x = 0) = 0.00000107374
P(x = 1) = ¹⁵C₁ 0.6¹ 0.4⁽¹⁵ ⁻ ¹⁾
P(x = 1) = 0.00002415919
Thus;
P( x > 1) = 1 - ( 0.00002415919 + 0.00000107374)
P( x > 1) = 0.999975
Probability that more than one student contracts the virus =0.999975
b.) The probability that more that 2 student contracts the virus is;
P(x > 2) = 1 - (P(x = 2) + P(x = 1) + P(x = 0))
P(x = 0) = 0.00000107374
P(x = 1) = 0.00002415919
P(x = 2) = 0.00025367151
Thus;
P(x > 2) = 1 - (0.00000107374 + 0.00002415919 + 0.00025367151
P(x > 2) = 0.999721
Thus, probability that more than 2 students contract the virus is:
P(x > 2) = 0.999721
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