Question

In a race, the odds in favour of the four horses H1, H2, H3, H4 are 1 : 4, 1 : 5, 1 : 6, 1 : 7, respectively.

Assuming that a dead heat is not possible, find the chance that one of them wins the race.​

Answer 1

Hi,

Answer:

The chance that one of them wins the race is 319/420.

Step-by-step explanation:

Given data:

The odds favoring the four horses are:

H1 = ¼

H2 = 1/5

H3 = 1/6

H4 = 1/7

Dead-heat is not possible i.e., tie between the horses is not possible. Anyone of them will win.

To find: The probability of any one of them winning

Let probability of the horses H1, H2, H3 & H4 winning be P(H1), P(H2), P(H3) & P(H4) respectively.

∴ Probability/chance of winning of (H1 or H2 or H3 or H4)  

= P(H1) + P(H2) + P(H3) + P(H4)  

= $\frac{1}{4}$ + $\frac{1}{5}$ + $\frac{1}{6}$ + $\frac{1}{7}$

= $\frac{105 + 84 + 70 + 60}{420}$

= $\frac{319}{420}$

Hope this helps!!!!!!

Answer 2

Answer:

319/420

Step-by-step explanation:

Let the responsibility of winning of the horses A = P(A)

Let the responsibility of winning of the horses B = P(B)

Let the responsibility of winning of the horses C = P(C)

Let the responsibility of winning of the horses D = P(D)

Therefore,

P(A) = 1/4,

P(B) = 1/5,

P(C) = 1/6,

P(D) = 1/7.

The events given are mutually exclusive, thus the chance that one of them wins  will be -

= P(AUBUCUD) = P(A) + P(B) + P(C) + P(D)

= (1/4) + (1/5) + (1/6) + (1/7).

= 319/420

Therefore, the chance that one of the horse wins the race is 319/420.​