Answer:
0.7361
Step-by-step explanation:
In this question we have
number to be 10
Then we have a probability of 10% = 0.10
We have q = 1-p
= 1-0.10 = 0.90
Then the probability of not more than 1 being defective:
P(x=0) + p(x= 1)
(10C0 x 0.1⁰ x 0.9^10-0)+(10C1 x 0.1¹ x 0.9^10-1)
= 1 x1 x0.3487 + 10 x 0.1 x 0.3874
= 0.3487 + 0.3874
= 0.7361
This is the the required probability and this answers the question.
probability = 10 percent = 0.1
q= 1- 10percent = 90% = 0.9
n = 4
To get the required probabiltiy for this question is
P(not greater than one is defective )=P(x=0)+P(x=1)
= 4C0x(0.1)⁰x(0.9)⁴+4C1x(0.1)¹x(0.9)³
= 0.9477
The required probability is 0.9477
11, 3, -3, -12
Hope this helps :)
Answer:
yes
Step-by-step explanation:
1+3=4
4+3=7
7+3=10
and 10+3=13
Answer:
uhhh ok
Step-by-step explanation:
