2/8 or 1/4 or 25 %
If both tosses are fair, the probability of that outcome is one in four.
It is 0.375It is 0.375It is 0.375It is 0.375
This is easiest to solve by working out the probability that no heads show and subtracting this from 1 to give the probability that at least one head shows: Assuming unbiased coins which won't land and stay on their edge, the probability of head = probability of tail = ½ → probability no heads = probability 5 tails = ½^5 = 1/32 → probability of at least one head = 1 - 1/32 = 31/32 = 0.96875 = 96.875 % = 96 7/8 %
1 in two but they say the side with heads is slightly Heavier.
2/8 or 1/4 or 25 %
(0.5)n
2/9
If both tosses are fair, the probability of that outcome is one in four.
2 out of 8
It is 0.375It is 0.375It is 0.375It is 0.375
This is easiest to solve by working out the probability that no heads show and subtracting this from 1 to give the probability that at least one head shows: Assuming unbiased coins which won't land and stay on their edge, the probability of head = probability of tail = ½ → probability no heads = probability 5 tails = ½^5 = 1/32 → probability of at least one head = 1 - 1/32 = 31/32 = 0.96875 = 96.875 % = 96 7/8 %
Theoretical probability = 0.5 Experimental probability = 20% more = 0.6 In 50 tosses, that would imply 30 heads.
1 in two but they say the side with heads is slightly Heavier.
It is 1/12.
None, since that would imply that in 18 cases the coin did not show heads or tails!
I assume you mean what's the chance of at least two heads showing when three fair coins are tossed. There are 8 possible outcomes as each coin can either be head or tails. For 3 heads, all 3 coins must show a head → 1 success For 2 heads, one coin will be a Tail; each coin could be a tail in turn → 3 successes → Pr = (1+3)/8 = 4/8 = 1/2 If you are wanting the probability that the first TWO specific coins are heads and the last, third, coin is either, then: Pr(head) = 1/2 → Pr(1st 2 heads, 3rd anything) = 1/2 × 1/2 × 1 = 1/4