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
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