A die is biased in such a way that if thrown once, then

Solution

Let the probability of getting an even number = 'x'
Then, probability of getting an odd number = {x + (1/5)}
So, x + {x + (1/5)} = 2x + (1/5) = 1
Or, x = {1 - (1/5)} ÷ 2 = (2/5)
So, probability of getting an even number and an odd number on the die are (2/5) and (3/5), respectively
Since there are 3 odd numbers and 3 even numbers on the die,
Probability of getting a specific even number = (2/5) × (1/3) = (2/15)
Similarly, probability of getting a specific odd number = (3/5) × (1/3) = (1/5)
On throwing two dice at once, all possible events where the sum is greater than or equal to 10 are
={(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)}
So, required probability = 3 × (2/15) × (2/15) + 2 × (1/5) × (2/15) + (1/5) × (1/5)
= 3 × (4/225) + 2 × (2/75) + 1 × (1/25)
= 12/225 + 6/225 + 9/225 = 27/225 = 11/75