In a carnival game, there are six identical boxes, one of which contains a prize. A contestant wins the prize by selecting the box containing it. Before each game, the old prize is removed and another prize is placed at random in one of the six boxes. Is it appropriate to use the binomial probability distribution to find the probability that a contestant who plays the game five times wins exactly twice? A) Yes. The five trials are independent, have only two outcomes, and have the same P(success); n = 5, r = 2, p = 1/5.
B) Yes. The five trials are independent, have only two outcomes, and have the same P(success); n = 2, r = 5, p = 1/6.
C) No. The five trials are independent, but have more than two outcomes.
D) Yes. The five trials are independent, have only two outcomes, and have the same P(success); n = 5, r = 2, p = 1/6.

D) Yes. The five trials are independent, have only two outcomes, and have the same P(success); n = 5, r = 2, p = 1/6.

Step-by-step explanation:

The Number of boxes = 6

Box containing a prize = 1

Probability of success, p = box containing a price / number of boxes = 1 /6

Number of trials = 5

Probability of success on exactly 2 trials, r = 2

Hence,

P(r = 2) = nCr * p^r * (1-p)^(n-r)

n = 5 ; r = 2 ; p = 1/6

Using a binomial probability calculator :

P(r = 2) = 0.1608

congruent segments

step-by-step explanation:

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in and ;

, as these are radii of our given circle.

therefore, by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

now let us substitute x=17 in the expression for the length of be.

therefore, length of be will be 27 units and option c is the correct choice.