A coin is flipped four times. X is the number of times it will land with a head up. Which statement is true about this situation? The variable X has a binomial distribution. P(success)=0.5P(success)=0.5; number trials = 4 The variable X does not have a binomial distribution because P(success) is not constant. The variable X does not have a binomial distribution because there are more than two possible outcomes. The variable X has a binomial distribution. P(success)=0.25; number of trials = 4
hint: what is the probability of heads on ONE coin?
0.5?
good so P(success) = 0.5 since there are two outcomes, this is a binomial distribution with number of trials = 4 so option A must be your answer
can you help with more please?
ok
For a certain type of hay fever, Medicine H has a 30% probability of working. In which distributions does the variable X have a binomial distribution? Select each correct answer. When the medicine is tried with two patients, X is the number of doses each patient needs to take. When the medicine is tried with six patients, X is the number of patients for whom the medicine does not work. When the medicine is tried with two patients, X is the number of patients for whom the medicine worked. When the medicine is tried with six patients, X is the number of patients for whom the medicine worked.
I believe it would be everything except the first option since X = the number of patients (success or fail)
if that doesn't work then I would only pick the last two
hint: if P(success) * P(success) = 0.36 then what is P(success)?
9?
no wait
what is the square root of 0.36?
0.6
good, that's your answer
hint: what is the probability of landing on A or B?
1/3
in probability "or" implies addition probability of A + probability of B = ?
2/3?
good, and if we square that (for X = 2) we get?
B?
what is (2/3)*(2/3) ?
1
what is 2*2?
4
good, what is 3*3?
9
good, so (2/3)*(2/3) = 4/9 so if X = 2 gives us a probability of 4/9 that means only option #1 (top left) works
The probability that traffic lights on a certain road will be green is 0.6. When a driver on that road approaches two traffic lights in a row, X is the number of traffic lights that are green. What is P(X=1)? Enter your answer, as a decimal,
if 0.6 is the probability of green, what is the probability of NOT green?
0.4?
good so we have two possibilities green * not green not green* green so that gives us (0.6)(0.4) + (0.4)(0.6) = ?
0.48
good that should be your answer
thanks!
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