A person forgets the last number of the phone number. He dials randomly and asks for the probability that he can dial correctly no more than three times? A .1/10 B .1/8 C.3/10 You can use the formula of 9 / 10 * 8 / 9 * 1 / 8 to calculate that the final answer is equal to answer A. I know, but what bothers me very much is that if you add the question no more than three times, then if you add the question no more than nine times, then the probability is still 1 / 10 when you dial nine times. It's a bit inexplicable here. Some of the translations haven't been practiced yet

A person forgets the last number of the phone number. He dials randomly and asks for the probability that he can dial correctly no more than three times? A .1/10 B .1/8 C.3/10 You can use the formula of 9 / 10 * 8 / 9 * 1 / 8 to calculate that the final answer is equal to answer A. I know, but what bothers me very much is that if you add the question no more than three times, then if you add the question no more than nine times, then the probability is still 1 / 10 when you dial nine times. It's a bit inexplicable here. Some of the translations haven't been practiced yet

Students, your answer is wrong. The answer here is 3 / 10. No more than three times can be correct. There may be: the first time is correct; the first time is wrong but the second time is correct; the first time and the second time are both wrong but the third time is correct. If the first time, the second time and the third time are both wrong and the fourth time is correct, it will be correct only after more than three times