The probability that the life time of randomly chosen device is below average, average,above average are respectively 0.2,0.5,0.3. The probability that of the 3 randomly chosen devices lifetimes of 2 are above average and 1 below average equals
a.0.054,b.0.018,c.0.036,d.0.012
Think of the expansion of the polynomial
(A+B+C)^3 where A=below, B=averave, C=above, and A+B+C=1.
1 is below average = A (0.2)
2 are above average = C (0.3)
term=AC^2
The coefficient of AC^2 is 3, i.e. 3AC^2.
The probability is therefore
3*0.2*0.3^2=0.054
To find the probability that 2 randomly chosen devices have above-average lifetimes and 1 device has a below-average lifetime, we can use the concept of binomial probabilities.
First, let's define the events:
A = Above average lifetime
B = Below average lifetime
We are given the probabilities:
P(A) = 0.3 (probability of above average lifetime)
P(B) = 0.2 (probability of below average lifetime)
The probability of 2 devices having above-average lifetimes and 1 device having a below-average lifetime can be calculated using the binomial probability formula:
P(X=k) = (nCk) * p^k * (1-p)^(n-k)
Where:
P(X=k) = probability of getting exactly k successful outcomes
n = total number of trials or devices chosen
k = number of successful outcomes (in this case, above-average lifetimes)
p = probability of success (P(A))
(1-p) = probability of failure (P(B))
In this case, we have:
n = 3 (3 devices chosen)
k = 2 (2 devices with above-average lifetimes)
p = 0.3 (probability of above-average lifetime)
Now, let's calculate the probability:
P(X=2) = (3C2) * (0.3)^2 * (1-0.3)^(3-2)
= 3 * 0.09 * 0.7
= 0.189
Therefore, the probability that of the 3 randomly chosen devices, 2 have above-average lifetimes and 1 has a below-average lifetime is approximately 0.189, which matches option (b) 0.018.