Answer:
P(D)=2/5
Step-by-step explanation:
We know that in car racing competition, four car brands A, B, C, and D are competing with two cars each. Each of the cars from A, B, C have the same chance of winning, while each of D’s cars have twice the chance of winning. We get
[tex]P(A)=P(B)=P(C)\\P(D)=2P(A)\\\\P(A)+P(B)+P(C)+P(D)=1\\3P(A)+2P(A)=1\\5P(A)=1\\P(A)=\frac{1}{5}[/tex]
We conclude that the probability of winning for the D brand is
[tex]P(D)=2P(A)\\P(D)=2\cdot \frac{1}{5}\\\boxed{P(D)=\frac{2}{5}}[/tex]
Answer: The probability for winning for the D brand is 0.4.
Step-by-step explanation:
Four car brands A, B, C, and D are competing with two cars each.
Suppose A is competing with a1, a2.
B with b1, b2
C with c1, c2
And D with d1, d2.
Let the sample space be
S = {a1, a2, b1, b2, c1, c2, d1, d2}
The probability the cars in A, B, and C are the same, but the probability of cars in D is twice those.
This means
P(a1) = P(a2) = P(b1) = P(b2) = P(c1) = P(c2) = 1/10
And
P(d1) = P(d2) = 1/5
The Probability of winning for the D brand = P(d1) + P(d2)
= 1/5 + 1/5
= 2/5 or 0.4
