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
The probability that a coin will land on heads - at least once - in six tosses is 0.9844
50/50
1 - (1/2)5 = 31/32
The probability is 1 out of 5
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 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.
The probability of two tails on two tosses of a coin is 0.52, or 0.25.
50/50
The probability is 1/4
Judy tosses a coin 4 times. draw a tree diagram showing the possible outcomes.What is the probability of getting at least 2 tails?
1 - (1/2)5 = 31/32
The probability is 1 out of 5
The probablility of getting 2 tails in 4 tosses of a fair coin is most likely 50%, 2/4=1/2, or .50.
The number of total outcomes on 3 tosses for a coin is 2 3, or 8. Since only 1 outcome is H, H, H, the probability of heads on three consecutive tosses of a coin is 1/8.
The experimental probability of a coin landing on heads is 7/ 12. if the coin landed on tails 30 timefind the number of tosses?
33%