a) Options E, F, G and H are all correct.
The distribution of X can be modelled by a binomial distribution because
E. The trials are independent.
F.There are a fixed number of bids.
G. There is the same probability of success for each trial (bid).
H. The data are binary (you either win the bid or not, that is, success or failure).
b) The probability of winning exactly 1 bid = 0.4410
c) The probability that you win 1 bid or fewer = 0.7840
d) The probability that you win more than 1 bid = 0.2160
Step-by-step explanation:
Complete Question
You are bidding on three items available on an online shopping site. You think that for each bid you have a 30% chance of winning it, and the outcomes of the three bids are independent events. Let X denote the number of winning bids out of the three items you bid on.
a. Explain why the distribution of X can be modeled by the binomial distribution.
A. The trials are dependent.
B. The data are not binary.
C. Probabilities of success for trials (bids) differ for each trial
D. The number of bids varies.
E. The trials are independent
F.There are a fixed number of bids.
G. There is the same probability of success for each trial (bid)
H. The data are binary
b. The probability of winning exactly 1 bid is (Round to four decimal places as needed.)
c. Find the probability that you win 1 bid or fewer.
d. Find the probability that you win more than 1 bid.
Solution
a) A binomial experiment is one in which the probability of success doesn't change with every run or number of trials.
It usually consists of a fixed number of runs/trials with only two possible outcomes, a success or a failure. The outcome of each trial/run of a binomial experiment is independent of one another.
Hence, the correct options are
E. The trials are independent.
F.There are a fixed number of bids.
G. There is the same probability of success for each trial (bid).
H. The data are binary (you either win the bid or not, that is, success or failure).
b) Binomial distribution function is represented by
P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ
n = total number of sample spaces = total number of bids available = 3
x = Number of successes required = winning exactly 1
p = probability of success = probability of winning a bid = 30% = 0.30
q = probability of failure = probability of NOT winning a bid = 1 - p = 1 - 0.30 = 0.70
P(X = 1) = ³C₁ (0.3)¹ (0.7)³⁻¹ = 0.4410 to 4 d.p.
c) The probability that you win 1 bid or fewer.
n = total number of sample spaces = total number of bids available = 3
x = Number of successes required = winning 1 or fewer bid = ≤ 1 = 0, 1
p = probability of success = probability of winning a bid = 30% = 0.30
q = probability of failure = probability of NOT winning a bid = 1 - p = 1 - 0.30 = 0.70
P(X ≤ 1) = P(X = 0) + P(X = 1)
P(X = 0) = ³C₀ (0.3)⁰ (0.7)³⁻⁰ = 0.343
P(X = 1) = 0.441 (as calculated in (b) above)
P(X ≤ 1) = 0.343 + 0.441 = 0.784
d) Probability that you win more than 1 bid
n = total number of sample spaces = total number of bids available = 3
x = Number of successes required = winning more than 1 bid = >1 = 2, 3
p = probability of success = probability of winning a bid = 30% = 0.30
q = probability of failure = probability of NOT winning a bid = 1 - p = 1 - 0.30 = 0.70
P(X > 1) = P(X = 2) + P(X = 3)
P(X = 2) = ³C₂ (0.3)² (0.7)³⁻² = 0.189
P(X = 3) = ³C₃ (0.3)³ (0.7)³⁻³ = 0.027
P(X > 1) = 0.189 + 0.027 = 0.2160
Hope this Helps!!!
