Question

a carnival game consists of 3 dart throws at a target. The probability of scoring a hit on any one throw is 30%. Using the binomial formula, find the probability of scoring 2 hits.

Answer 1

Answer:  Probability of scoring 2 hits = 0.63.

Step-by-step explanation:

Since we have given that

Number of dart throws at a target = 3

Probability of scoring a hit on any one throw = 30%

We will use "Binomial Distribution" i.e.

$P=^nC_rp^r(1-p)^{n-r}$

where,

n denotes number of dart throws at a target,

r denotes number of required throws

p denotes probability of success

(1-p) denotes probability of failure

So, Probability of success is given by

$\frac{30}{100}=\frac{3}{10}$

Probability of failure is given by

$1-\frac{30}{100}=\frac{70}{100}=\frac{7}{10}$

We will use "Binomial Distribution" i.e.

$P(X=2)=^3C_2(\frac{3}{10})^2\times (\frac{7}{10})\\\\P(X=2)=\frac{9}{100}\times \frac{7}{100}\\\\P(X=2)=\frac{63}{100}\\\\P(X=2)=0.63$

Hence, Probability of scoring 2 hits = 0.63.