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
The probability is 6/36 = 1/6.
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
The probability is 6/36 = 1/6.
1/4 if they are tossed only once.
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
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} ).