50/50
The probability that a coin will land on heads - at least once - in six tosses is 0.9844
Pr(At least one head in three tosses) = 1 - Pr(No heads in three tosses) = 1 - Pr(Three tails in three tosses) = 1 - (1/2)*(1/2)*(1/2) = 1 - 1/8 = 7/8 or 0.875 or 87.5%
Pr(At least one head in 3 tosses) = 1 - Pr(No heads in 3 tosses) = 1 - Pr(3 tails in three tosses) = 1 - [Pr(T)*Pr(T)*Pr(T)] since the three tosses are independent. = 1 - 1/2 * 1/2 *1/2 = 1 - 1/8 = 7/8
Pr(H at least once in 10 tosses) = 1 - pr(No H in 10 tosses) = 1 - Pr(10 T in 10 tosses) = 1 - (1/2)10 = 1 - 1/1024 = 1023/1024
50/50
The probability that a coin will land on heads - at least once - in six tosses is 0.9844
The probability of tossing heads on all of the first six tosses of a fair coin is 0.56, or 0.015625. The probability of tossing heads on at least one of the first six tosses of a fair coin is 1 - 0.56, or 0.984375.
Pr(At least one head in three tosses) = 1 - Pr(No heads in three tosses) = 1 - Pr(Three tails in three tosses) = 1 - (1/2)*(1/2)*(1/2) = 1 - 1/8 = 7/8 or 0.875 or 87.5%
Pr(At least one head in 3 tosses) = 1 - Pr(No heads in 3 tosses) = 1 - Pr(3 tails in three tosses) = 1 - [Pr(T)*Pr(T)*Pr(T)] since the three tosses are independent. = 1 - 1/2 * 1/2 *1/2 = 1 - 1/8 = 7/8
Pr(H at least once in 10 tosses) = 1 - pr(No H in 10 tosses) = 1 - Pr(10 T in 10 tosses) = 1 - (1/2)10 = 1 - 1/1024 = 1023/1024
50/50
The probablility of getting 2 tails in 4 tosses of a fair coin is most likely 50%, 2/4=1/2, or .50.
The probability of tossing 6 heads in 6 dice is 1 in 26, or 1 in 64, or 0.015625. THe probability of doing that at least once in six trials, then, is 6 in 26, or 6 in 64, or 3 in 32, or 0.09375.
To find P( at least 1 head in 7 tosses) we can find P( no heads) and subtract that from one. Alternatively, we need to find P ( 1 head) + P ( 2 heads )+...+ P(7 heads) Since P of no heads is P( all tails) this is (1/2)7 =.0078125 ( 1/128 as a fraction) Now 1-(1/128)=127/128 or .9921875
1 - (1/2)5 = 31/32
the only combination which does not produce heads at least once it tails twice. the odds of getting tails twice is 0.5*0.5=0.25 so the odds for getting heads at least once is 1-0.25=0.75 or 75% or 3/4.
This can be calculated easily by multiplying the chances of getting one head in one toss for a fair coin (half) by itself(half) to give a quarter (1/4 or 1 in 4). If you wanted the chances of at least one head in 2 tosses, then the chances would be for anyhting exept 2 tails in a row, which is 1 - 0.25(quarter), which is 0.75 (3/4 or 3 quarters or 3 in 4)