2 out of 6
To find the probability of getting at least one head in 4 coin tosses, it's easier to calculate the complementary probability of getting no heads at all (i.e., getting all tails). The probability of getting tails in a single toss is 0.5, so for 4 tosses, the probability of all tails is ( (0.5)^4 = 0.0625 ). Therefore, the probability of getting at least one head is ( 1 - 0.0625 = 0.9375 ) or 93.75%.
In a large enough number of tosses, it is a certainty (probability = 1). In only the first three tosses, it is (0.5)3 = 0.125
The probability is 1 out of 5
In a series of ten coin tosses, each toss has two possible outcomes: heads or tails. The expected number of heads can be calculated as the product of the number of tosses and the probability of getting heads in a single toss, which is 0.5. Therefore, in ten tosses, the expected number of heads is 10 × 0.5 = 5 heads. However, the actual number of heads can vary due to the randomness of each toss.
The probability is 0, since there will be some 3-tosses in which you get 0, 1 or 3 heads. So not all 3-tosses will give 2 heads.
To find the probability of getting at least one head in 4 coin tosses, it's easier to calculate the complementary probability of getting no heads at all (i.e., getting all tails). The probability of getting tails in a single toss is 0.5, so for 4 tosses, the probability of all tails is ( (0.5)^4 = 0.0625 ). Therefore, the probability of getting at least one head is ( 1 - 0.0625 = 0.9375 ) or 93.75%.
In a large enough number of tosses, it is a certainty (probability = 1). In only the first three tosses, it is (0.5)3 = 0.125
The probability of two tails on two tosses of a coin is 0.52, or 0.25.
The probability is 0.0322
The probability is 1/4
The probability is 1 out of 5
0.5
In a series of ten coin tosses, each toss has two possible outcomes: heads or tails. The expected number of heads can be calculated as the product of the number of tosses and the probability of getting heads in a single toss, which is 0.5. Therefore, in ten tosses, the expected number of heads is 10 × 0.5 = 5 heads. However, the actual number of heads can vary due to the randomness of each toss.
If the coin is fair, the probability of getting all heads will decrease exponentially towards 0.
The probability is 0, since there will be some 3-tosses in which you get 0, 1 or 3 heads. So not all 3-tosses will give 2 heads.
1/4
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} ).