The chance of India winning toss is `3//4.` If it wins the toss, then its chance of victory is `4//5` otherwise it is only `1//2.` Then chance of India's vectory is

Answer

To find the chance of India winning the match, we can break down the problem into two cases based on whether India wins the toss or loses the toss.

Let's denote:
- \( P(T) \): Probability of India winning the toss = \( \frac{3}{4} \)
- \( P(W|T) \): Probability of India winning the match given that it wins the toss = \( \frac{4}{5} \)
- \( P(L) \): Probability of India losing the toss = \( 1 - P(T) = 1 - \frac{3}{4} = \frac{1}{4} \)
- \( P(W|L) \): Probability of India winning the match given that it loses the toss = \( \frac{1}{2} \)

Now, we can use the law of total probability to find the overall probability of India winning the match \( P(W) \):

\[ P(W) = P(T) \cdot P(W|T) + P(L) \cdot P(W|L) \]

Substituting the values we have:

\[ P(W) = \left(\frac{3}{4}\right) \cdot \left(\frac{4}{5}\right) + \left(\frac{1}{4}\right) \cdot \left(\frac{1}{2}\right) \]

Calculating each term:

1. For the first term:
\[ \frac{3}{4} \cdot \frac{4}{5} = \frac{12}{20} = \frac{3}{5} \]

2. For the second term:
\[ \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8} \]

Now we need to add these two results together:
To add \( \frac{3}{5} \) and \( \frac{1}{8} \), we need a common denominator. The least common multiple of 5 and 8 is 40.

Converting each fraction:
\[ \frac{3}{5} = \frac{3 \times 8}{5 \times 8} = \frac{24}{40} \]
\[ \frac{1}{8} = \frac{1 \times 5}{8 \times 5} = \frac{5}{40} \]

Now we can add them:
\[ P(W) = \frac{24}{40} + \frac{5}{40} = \frac{29}{40} \]

Thus, the chance of India winning the match is:
\[ \frac{29}{40} \]

### Final Answer:
The chance of India's victory is \( \frac{29}{40} \).