To solve this problem, we make use of the Binomial
Probability equation which is mathematically expressed as:
P = [n! / r! (n – r)!] p^r * q^(n – r)
where,
n = the total number of gadgets = 4
r = number of samples = 1 and 2 (since not more than 2)
p = probability of success of getting a defective gadget
q = probability of failure = 1 – p
Calculating for p:
p = 5 / 15 = 0.33
So,
q = 1 – 0.33 = 0.67
Calculating for P when r = 1:
P (r = 1) = [4! / 1! 3!] 0.33^1 * 0.67^3
P (r = 1) = 0.3970
Calculating for P when r = 2:
P (r = 2) = [4! / 2! 2!] 0.33^2 * 0.67^2
P (r = 2) = 0.2933
Therefore the total probability of not getting more than
2 defective gadgets is:
P = 0.3970 + 0.2933
P = 0.6903
Hence there is a 0.6903 chance or 69.03% probability of
not getting more than 2 defective gadgets.
Answer:
He could multiply the amount one box ways by 100, or 0.6*100.
Step-by-step explanation:
