when two dice are tossed , 6*6 = 36 outcomes are possible
a max output of 4 as defined can occur through
1-4 , 2-4 , 3-4 , 4-4 , 4-3 , 4-2 , & 4-1 i.e. 7 ways
so its probability is 7/36
the problem thus reduces to a binomial distribution
n = number of "trials" = 30
p = probability of "success" on a single trial = 7/36
q = 1 - p = probability of "failure" on a single trial = 29/36
the formula for k successes in n trials is given by
P[k] = nCk*p^k*q^(n-k)
thus reqd. probability = P[16] + P[17]
= 30C16*[7/36]^16*[29/36]^14 + 30C17*[7/36]^17*[29/36]^13
= 3.527/10^5
Coins do not have numbers, there is only the probability of heads or tails.
The probability that the die tossed will land on a number that is smaller than 5 is 4/6 or 2/3. Smaller than 5 is 1 - 4 and 6 is the sample space.
The probability is 1/16.
1/6
1/4 if they are tossed only once.
Coins do not have numbers, there is only the probability of heads or tails.
The probability that the die tossed will land on a number that is smaller than 5 is 4/6 or 2/3. Smaller than 5 is 1 - 4 and 6 is the sample space.
The probability is 90/216 = 5/12
A single fair die has the numbers 1 to 6, so when a single fair die is tossed the probability of obtaining a number different than 11 is: P(x diff than11) = 1.
The probability is 1/16.
1/6
1/4 if they are tossed only once.
The probability is 6/36 = 1/6.
The probability is 50-50.
The number of times a coin is tossed does not alter the probability of getting heads, which is 50% in every case, as long as the coin has not been rigged (i.e., a double-headed coin, a weighted coin) to alter the result.
The probability is 0.5The probability is 0.5The probability is 0.5The probability is 0.5
It is 0.5