To find the probability that a five-digit number has at least one zero in it, we can use the complementary probability approach. Let's go through the solution step-by-step:
First, calculate the total number of five-digit numbers. A five-digit number ranges from 10000 to 99999. Therefore, the total number of five-digit numbers is 99999 - 10000 + 1 = 90000.
Next, determine the number of five-digit numbers that do not contain any zeros. For a number to have no zeros:
Therefore, the total number of five-digit numbers with no zeros is 9^5. Calculating this gives:
9^5 = 9 \times 9 \times 9 \times 9 \times 9 = 59049Now, use the complementary probability to find the number of five-digit numbers with at least one zero:
90000 - 59049 = 30951
Finally, calculate the probability that a five-digit number has at least one zero:
Probability = \frac{30951}{90000}
Simplifying the fraction gives approximately 0.344.
Therefore, the probability that a five-digit number has at least one zero in it is 0.344.
If the probability distribution is given by:
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|---|
| P(x) | 0 | k | 2k | 2k | 3k | k² | 2k² | 7k² + k |
Then find: \( P(3 < x \leq 6) \)
If \(S=\{1,2,....,50\}\), two numbers \(\alpha\) and \(\beta\) are selected at random find the probability that product is divisible by 3 :
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A shopkeeper marks his goods 40% above cost price and offers a 10% discount. What is his percentage profit?