Answer: 0.7209
Step-by-step explanation:
For a fair die, probability of occurrence of any face (including 5) = 1/6.
To determine the probability of having at least one *5" in 7 rolls, the probability distribution formula for occurrence of random variables by combination method is used.
To determine no of occurrence "r" from a total occurrence "n", the probability formula is denoted by:
P(X=r) = nCr × p^r × q^n-r
Where p= probability of success which in this case = 1/6
q = 1-p = 1 - 1/6 = 5/6
To determine the probability of the occurrence of at least one occurrence of "5" from 7 throws, that is P(X≥1), we can determine the probability of occurrence of not having 5 at all and deduct our answer from 1. That is, P(X≥1) = 1 - P(X=0)
When r = 0,
P(X=0) = 7C0 × (1/6)^0 × (5/6)^7 = 0.2791
Hence, P(X≥1) = 1 - 0.2791 = 0.7209
The required probability is[tex]P(x\ge 1)=0.72092[/tex]
The formula for probability mass function is,
[tex]P(X=x)=\frac{n!}{x!(n-x)!} p^x(1-p)^{n-x}[/tex]
It is given that the probability of getting 5 when a fair die is rolled is [tex]\frac{1}{6}[/tex]
A fair die is thrown [tex]n=7[/tex] times
Let [tex]x[/tex] be the number of times 5 comes up
So, the pmf of [tex]x[/tex] is,
[tex]P(x\ge 1)=1-P(X=0)\\=1-\frac{7!}{0!(7-0)!} (\frac{1}{6} )^0(\frac{5}{6} )^7\\=1-\frac{78125}{279936}\\ =1-0.27908\\P(x\ge 1)=0.72092[/tex]
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