Percentage of cases they will contradict where atleast one of them is correct

Braden and Fred are two independent journalists. Braden is usually correct in 33% of his reports and Fred in 70% of his reports. In what percentage of cases are they likely to contradict each other, talking about the same incident where at least one of them is correct.

Is this question supposed to solved with probability?

This is indeed supposed to be solved with probability.

There are 4 cases

• Braden is right and Fred is right
• Braden is right and Fred is wrong
• Braden is wrong and Fred is right
• Braden is wrong and Fred is wrong

We also know that the probability of two unrelated events both happening is equal to the product of the probabilities of the events:

From this, we can see that the probability of the first case is equal to the probability that Braden is right, multiplied by the probability that Fred is right

Using this logic on all four cases, we get the following probabilities:

Since we are only interested in the cases where they contradict each other, we can simply sum the probability of those two cases.

From that, we can see that the probability of the two contradicting each other is 17/30, or 57%