Let x be the number of defective cell phones. It is given that in batch of 100, on average 5 are defective.
let p be the probability of defective cell phone.
p = 5/100 = 0.05
Let n be size of random sample, n=30
Here out of 30 we want to find probability that 2 will be defective. It means 30-2 =28 cell phones will be non defective.
The probability of getting non defective cell phone is 1- p=1-0.05 =0.95
The probabability of getting 2 defective is
P(X=2) = number of ways selecting 2 from 30 * probability 2 defective * probability of 28 non defective
Now number of ways of selecting 2 cell phone from 30 is
30C2 = 
= 
= 
= (30*29) /2
30C2 = 435
P(X=2) = 30C2 * (0.05)^2 * (0.95)^28
= 435 * 0.0025 * 0.2378
P(X=2) = 0.2586
Probability of getting 2 defective out of 30 is 0.2586
The values for y are as follows
4.5, 4, 3.5, 3, 2.5, 2
The possibility would be 50 percent if 3 is included.
Answer:
Multiply by 1.16
Step-by-step explanation:
Answer:
2/3
Step-by-step explanation:
fraction of book already read = 1/3
fraction which represents whole book = 1 = 3/3
fraction of remaining book = fraction of whole book - fraction of book already read
= 1 - 1/3
= 3/3 - 1/3
= (3-1) /3
= 2/3