Since the dice are fair and the rolling are independent, each single outcome has probability 1/15. Every time we choose
We have 
and 
, because the dice are fair.
Now we use the assumption of independence to claim that
Now, we simply have to count in how many ways we can obtain every possible outcome for the sum. Consider the attached table: we can see that we can obtain:
- 2 in a unique way (1+1)
- 3 in two possible ways (1+2, 2+1)
- 4 in three possible ways
- 5 in three possible ways
- 6 in three possible ways
- 7 in two possible ways
- 8 in a unique way
This implies that the probabilities of the outcomes of 
are the number of possible ways divided by 15: we can obtain 2 and 8 with probability 1/15, 3 and 7 with probability 2/15, and 4, 5 and 6 with probabilities 3/15=1/5
Answer:
c
Step-by-step explanation:
Let x be the random variable representing defective phone. Let n be the sample size. Let p be the probability that phone is defective.
Given: n=500, p= 0.02
From given information we know that x is random variable such that p is the probability of success and it is constant for each trial. Sample size n is fixed.
X follows Binomial distribution with parameters n=500 and p=0.02
a). The average number of defective phone
E(x) = n*p = 500 * 0.02 = 10
The average number of defective phones is 10.
b)Probability of getting 5 defective phones.
P(X=5) = 
= 
= 0.037
The probability of getting exactly 5 defective is 0.037.
