To find the probability of getting an odd number at least once when a die is tossed thrice, we can use the complementary approach. The probability of not getting an odd number (i.e., getting an even number) in a single toss is ( \frac{3}{6} = \frac{1}{2} ). Therefore, the probability of getting an even number in all three tosses is ( \left(\frac{1}{2}\right)^3 = \frac{1}{8} ). Thus, the probability of getting an odd number at least once is ( 1 - \frac{1}{8} = \frac{7}{8} ).
(1/2)^3 = 1/8th Since the initial probability (assuming independence) of getting a head in a single toss is one half (1/2) we just cube this probability because of the number of events we are performing. So if you were to try to calculate the probability of a coin being tossed 6 times it would be one half to the 6th power which is 1/64.
The probability is 50-50.
The probability is 1/4
Coins do not have numbers, there is only the probability of heads or tails.
50%
(1/2)^3 = 1/8th Since the initial probability (assuming independence) of getting a head in a single toss is one half (1/2) we just cube this probability because of the number of events we are performing. So if you were to try to calculate the probability of a coin being tossed 6 times it would be one half to the 6th power which is 1/64.
The probability is 50-50.
It is 0.25
The probability is 1/4
Coins do not have numbers, there is only the probability of heads or tails.
The probability is 3/8 = 0.375
The probability is 0.5The probability is 0.5The probability is 0.5The probability is 0.5
50%
1/2
possible
50%
It is 0.75