Respuesta :
Answer:
The probability of hitting the bullseye at least once in 6 attempts is 0.469.
Step-by-step explanation:
It is given that a mechanical dart thrower throws darts independently each time, with probability 10% of hitting the bullseye in each attempt.
The probability of hitting bullseye in each attempt, p = 0.10
The probability of not hitting bullseye in each attempt, q = 1-p = 1-0.10 = 0.90
Let x be the event of hitting the bullseye.
We need to find the probability of hitting the bullseye at least once in 6 attempts.
[tex]P(x\geq 1)=1-P(x=0)[/tex] .... (1)
According to binomial expression
[tex]P(x=r)=^nC_rp^rq^{n-r}[/tex]
where, n is total attempts, r is number of outcomes, p is probability of success and q is probability of failure.
The probability that the dart thrower not hits the bullseye in 6 attempts is
[tex]P(x=0)=^6C_0(0.10)^0(0.90)^{6-0}[/tex]
[tex]P(x=0)=0.531441[/tex]
Substitute the value of P(x=0) in (1).
[tex]P(x\geq 1)=1-0.531441[/tex]
[tex]P(x\geq 1)=0.468559[/tex]
[tex]P(x\geq 1)\approx 0.469[/tex]
Therefore the probability of hitting the bullseye at least once in 6 attempts is 0.469.
