A group of freshmen at a local university consider joining the equestrian team. Thirty‑five percent of students choose Western riding, 45% choose dressage, and 50% choose jumping. Twenty percent choose both dressage and jumping, while 10% choose Western and dressage. No one chooses Western and jumping. There are no horses suitable for two styles, and each student is assigned to one horse.
A - What is the probability that a student chooses dressage or jumping?a) 0.95b) 0.9c) 0.23d) 0.75
B - What is the probability that a student chooses neither dressage nor Western riding?a) 0.05b) 0.2c) 0.5d) 0.3

Respuesta :

Answer:

a) Option D) 0.75

b) Option D) 0.3

Step-by-step explanation:

We are given the following in the question:

Percentage of students who choose Western riding = 35%

[tex]P(w) = 0.35[/tex]

Percentage of students who choose dressage= 45%

[tex]P(d) = 0.45[/tex]

Percentage of students who choose jumping = 50%

[tex]P(j) = 0.50[/tex]

Percentage of students who choose both dressage and jumping = 20%

[tex]P(d \cap j) = 0.20[/tex]

Percentage of students who choose Western and dressage = 10%

[tex]P(w \cap d) = 0.10[/tex]

Percentage of students who choose Western and jumping = 0%

[tex]P(w \cap j) = 0[/tex]

Thus, we can say

[tex]P(w \cap j \cap d) = 0[/tex]

Formula:

[tex]P(A \cup B) = P(A) + P(B) - P(A \cap B)[/tex]

a) P(student chooses dressage or jumping)

[tex]P(d \cup j) = P(d) + P(j) -p(d \cap j)\\= 0.45 + 0.50-0.20 = 0.75[/tex]

b) P(student chooses neither dressage nor Western riding)

[tex]= 1 - P(d \cup w)\\= 1 - (P(d) + P(w) - P(d \cap w))\\= 1 - (0.45 + 0.35 - 0.10)= 0.3[/tex]

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