According to the question, it is stated that 24% of the students at the school that are involved in a sports team also participated in the prom dance.

24% = 0.24

This means that we are going to find 24% of the original 56%, since 24% of them also participated in the prom dance.

The probability that a student who is involved in a sports team also participated in the prom dance = 0.24 * 0.56

The probability that a student who is involved in a sports team also participated in the prom dance = 0.1344

5 0

3 years ago

The route used by a certain motorist in commuting to work contains two intersections with traffic signals. The probability that

tekilochka [14]

Answer:

a) P(X∩Y) = 0.2

b) $P_1$ = 0.16

c) P = 0.47

Step-by-step explanation:

Let's call X the event that the motorist must stop at the first signal and Y the event that the motorist must stop at the second signal.

So, P(X) = 0.36, P(Y) = 0.51 and P(X∪Y) = 0.67

Then, the probability P(X∩Y) that the motorist must stop at both signal can be calculated as:

P(X∩Y) = P(X) + P(Y) - P(X∪Y)

P(X∩Y) = 0.36 + 0.51 - 0.67

P(X∩Y) = 0.2

On the other hand, the probability $P_1$ that he must stop at the first signal but not at the second one can be calculated as:

$P_1$ = P(X) - P(X∩Y)

$P_1$ = 0.36 - 0.2 = 0.16

At the same way, the probability $P_2$ that he must stop at the second signal but not at the first one can be calculated as:

$P_2$ = P(Y) - P(X∩Y)

$P_2$ = 0.51 - 0.2 = 0.31

So, the probability that he must stop at exactly one signal is:

$P = P_1+P_2\\P=0.16+0.31\\P=0.47$

7 0

3 years ago