# A binomial experiment has 6 trials in which p = 0.85. What is the probability of getting at least 4 successes? A) 0.9527 September 26, 2023 by Thái Dương

A binomial experiment has 6 trials in which p = 0.85. What is the probability of getting at least 4 successes?
A) 0.9527
B) 0.9178
C)0.4554

### 2 thoughts on “A binomial experiment has 6 trials in which p = 0.85. What is the probability of getting at least 4 successes? A) 0.9527<br /”

0.9527
Step-by-step explanation:
Once again, if Brainly would stop deleting my answer for no reason, your answer is 0.9527. OKAY! …okay

2. The binomial probability for getting at least 4 successes is 0.955

### What is binomial probability?

The probability of exactly x successes on n repeated trials in an experiment which has two possible outcomes is called binomial probability.

### Binomial probability formula

$$P_{x} = nC_{x} p^{x} q^{n-x}$$
Where,
P is the binomial probability
x is the number of times for a specific outcomes within n trials
$$nC_{x}$$ is the number of combinations
p is the probability of success in a single trial
q is the probability of failure on a single trial
n is the number of trials
Let p denote the probability of getting success in a single trial and q denotes the probability of failure in a single trial.
And let x be a random variable denoting the number of success.
According to the given question.
We have
$$p = 0.85$$
⇒$$q = 1- 0.85 = 0.15$$
Also,
n = 6
Therefore,
The  binomial probability of getting at least 4 success
$$=P(x\geq 4)$$
$$=P(x=4)+P(x=5)+P(x=6)$$
$$= 6C_{4} (0.85)^{4} (0.15)^{2} + 6C_{5} (0.85)^{5} (0.15)^{1}+ 6C_{6} (0.85)^{6} (0.15)^{0}$$
$$=\frac{6!}{4!21} (0.522)(0.0225)+\frac{6!}{5!1!} (0.443)(0.15)+\frac{6!}{6!0!} (0.4)$$
$$=15(0.011)+6(0.066)+1(0.4)$$
$$=0.165+0.396+0.4\\=0.955$$
Hence, the binomial probability for getting at least 4 successes is 0.955
Thus, option A is correct.