Answer:
0.0256 ; 0.1536 ; 0.3456 ; 0.3456 ; 0.1296
Step-by-step explanation:
P, that each bulldozer will be operational for atleast 6 months = 0.6
p = 0.6 ; q = 1 - p = 1 - 0.6 = 0.4
Number of bulldozers purchased, n = 4
Probability that :
Exactly 0 bulldozer will be operational at the end of 6 months :
P(x = x) = nCx * p^x * q^(n-x)
P(x = 0) = 4C0 * 0.6^0 * 0.4^4
P(x = 0) = 1 * 1 * 0.0256
P(x = 0) = 0.0256
Probability that :
Exactly 1 bulldozer will be operational at the end of 6 months :
P(x = x) = nCx * p^x * q^(n-x)
P(x = 1) = 4C1 * 0.6^1 * 0.4^3
P(x = 1) = 4 * 0.6 * 0.064
P(x = 1) = 0.1536
Probability that :
Exactly 2 bulldozer will be operational at the end of 6 months :
P(x = x) = nCx * p^x * q^(n-x)
P(x = 2) = 4C2 * 0.6^2 * 0.4^2
P(x = 2) = 6 * 0.36 * 0.16
P(x = 2) = 0.3456
Probability that :
Exactly 3 bulldozer will be operational at the end of 6 months :
P(x = x) = nCx * p^x * q^(n-x)
P(x = 3) = 4C3 * 0.6^3 * 0.4^1
P(x = 3) = 4 * 0.216 * 0.4
P(x = 3) = 0.3456
Probability that :
Exactly 4 bulldozer will be operational at the end of 6 months :
P(x = x) = nCx * p^x * q^(n-x)
P(x = 4) = 4C4 * 0.6^4 * 0.4^0
P(x = 1) = 1 * 0.1296 * 1
P(x = 1) = 0.1296
