Question

Two fair dice are tossed independently and it is found that one face is odd and the other one is even. Then the probability (round off to 2 decimal places) that the sum is less than 6 equals .............

Hint:

When calculating probabilities with restrictions, first identify all possible favorable outcomes and then compute the ratio of favorable outcomes to the total number of outcomes.

Answer 1

Correct Answer: 0.3

Step 1: Understand the given condition.
We are given that one die shows an odd number and the other die shows an even number. The odd numbers on a die are \(1, 3, 5\), and the even numbers are \(2, 4, 6\).
Step 2: Count the total number of favorable outcomes.
Since one die shows an odd number and the other shows an even number, the possible pairs of outcomes are: \[ (1, 2), (1, 4), (1, 6), (3, 2), (3, 4), (3, 6), (5, 2), (5, 4), (5, 6) \] Thus, there are 9 favorable outcomes.
Step 3: Find the sum for each favorable outcome.
The sum of the pairs is: \[ 1 + 2 = 3, \quad 1 + 4 = 5, \quad 1 + 6 = 7, \quad 3 + 2 = 5, \quad 3 + 4 = 7, \quad 3 + 6 = 9, \quad 5 + 2 = 7, \quad 5 + 4 = 9, \quad 5 + 6 = 11 \] Out of these, the sums less than 6 are \(3\) and \(5\). These correspond to the pairs: \[ (1, 2), (1, 4), (3, 2), (5, 2). \] Thus, there are 4 favorable outcomes where the sum is less than 6.
Step 4: Calculate the probability.
The probability is the ratio of favorable outcomes to total outcomes. There are \(6 \times 3 = 18\) possible outcomes where one die is odd and the other is even (since there are 3 odd numbers and 3 even numbers). Therefore, the probability is: \[ \frac{4}{18} = \frac{2}{9} \approx 0.22. \]
Final Answer: \[ \boxed{0.22}. \]