Answer:
a.
b. 
c. 
d. 
Step-by-step explanation:
The probability that the company won x bids follows a binomial distribution because we have n identical and independent experiments with a probability p of success and (1-p) of fail.
So, the PMF of X is equal to:

Where p is 0.1 and it is the chance of winning. Additionally, n is 3 and it is the number of bids. So the PMF of X is:

For binomial distribution:

Therefore, the company can expect to win 0.3 bids and it is calculated as:

Additionally, the standard deviation of the number of bids won is:

Finally, the probability to won 1, 2 or 3 bids is equal to:

So, the expected profit for the company is equal to:

Because there is a probability of 0.243 to win one bid and it will produce 50,000 of income, there is a probability of 0.027 to win 2 bids and it will produce 100,000 of income and there is a probability of 0.001 to win 3 bids and it will produce 150,000 of income.