Let \(O\) be the event of getting an odd number when a die is thrown.
Let \(E\) be the event of getting an even number when a die is thrown.
The possible outcomes when a die is thrown are {1, 2, 3, 4, 5, 6}.
The odd numbers are {1, 3, 5}.
The even numbers are {2, 4, 6}.
The probability of getting an odd number is \(P(O) = \frac{3}{6} = \frac{1}{2}\).
The probability of getting an even number is \(P(E) = \frac{3}{6} = \frac{1}{2}\).
The die is thrown 10 times. We are looking for the probability that an odd number comes up at least once. It's easier to calculate the probability of the complement event (that no odd numbers come up, meaning only even numbers appear) and subtract it from 1.
The probability of getting an even number in one throw is \(P(E) = \frac{1}{2}\).
The probability of getting even numbers in all 10 throws is \(P(E)^{10} = \left(\frac{1}{2}\right)^{10} = \frac{1}{1024}\).
The probability of getting at least one odd number is \(1 - P(\text{no odd numbers}) = 1 - \frac{1}{1024} = \frac{1024 - 1}{1024} = \frac{1023}{1024}\).
Thus, the probability that an odd number will come up at least one time is \(\frac{1023}{1024}\).
The probability of getting an odd number in a single throw is:
$$ P(\text{odd}) = \frac{3}{6} = \frac{1}{2}. $$
The probability of not getting an odd number (i.e., getting an even number) in a single throw is:
$$ P(\text{even}) = 1 - P(\text{odd}) = 1 - \frac{1}{2} = \frac{1}{2}. $$
The probability of not getting an odd number in all 10 throws is:
$$ P(\text{no odd in 10 throws}) = \left(\frac{1}{2}\right)^{10} = \frac{1}{1024}. $$
The probability of getting an odd number at least once is the complement of the above:
$$ P(\text{at least one odd}) = 1 - P(\text{no odd in 10 throws}) = 1 - \frac{1}{1024} = \frac{1024 - 1}{1024} = \frac{1023}{1024}. $$
Final Answer: The final answer is $ {\frac{1023}{1024}} $.
A wooden block of mass M lies on a rough floor. Another wooden block of the same mass is hanging from the point O through strings as shown in the figure. To achieve equilibrium, the coefficient of static friction between the block on the floor and the floor itself is
In an experiment to determine the figure of merit of a galvanometer by half deflection method, a student constructed the following circuit. He applied a resistance of \( 520 \, \Omega \) in \( R \). When \( K_1 \) is closed and \( K_2 \) is open, the deflection observed in the galvanometer is 20 div. When \( K_1 \) is also closed and a resistance of \( 90 \, \Omega \) is removed in \( S \), the deflection becomes 13 div. The resistance of galvanometer is nearly: