In single die,
S = {1,2,3,4,5,6} So n(S) = 6
The event A is defined as getting a number less than 3.
A = {1,2} So n(A)=2
Hence P(A) = n(A)/n(S) = 2/6 =1/3
The probability is 0.0322
33%
The experimental probability of a coin landing on heads is 7/ 12. if the coin landed on tails 30 timefind the number of tosses?
75%. There are 3 possible ways of getting at least one tail from 2 tosses from a coin:Tail & Tail orHead & Tail orTail & HeadEach of these individual outcomes has a probability of 25% (e.g. the probability of getting a tail and then another tail is 25%). Adding the possible outcomes together gives you a total of 75%.
1/2 or 5/10
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 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
If the coin is fair, the probability of getting all heads will decrease exponentially towards 0.
1/4
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 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} ).