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.
To find the answer we can find the amount he will get paid separately.
The amount he will get paid on site A:
0.75 x 5
=$3.75
The amount he will get paid on site B:
0.25 x 5
= $1.25
To find the amount Joe get paid at site A than at site B, we can deduct the amount he get from site B by the amount he get from site A:
= 3.75 - 1.25
= $2.50
Therefore $2.50 is the answer.
Hope it helps!
(-75/37, -193/74) is your answer .-.
Answer:
In my opinion, the holding pen (writing or solving something) and being at war related to 3 phases.
Phase 0: Be serious with your pen, with your work.
Phase 1 : After making decision, start to draft carefully, you should list down everything which could happen, prepare well.
Phase 2: Enter the process, enter the battle, keep calm and be decisive.
< Make it quick, I guess this would be deleted as soon>
