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What is the probability of heads heads tails tails heads heads on six flips of a coin?
(1/2)*(1/2)*(1/2)*(1/2)*(1/2)*(1/2) = 1/26 = 1/64
What is the most heads a cat has been born with?
2
What is the probablity that the sequence will be heads-heads-heads-heads if you toss a coin 4 times?
Each independent trial has a 1/2 probability that a heads will result. So for a sequence of 4 heads you have 1/2 x 1/2 x 1/2 x 1/2 = 1/16
What is the probability of tossing a coin three times and getting it on heads each time?
Each toss has a 1/2 probability of getting heads. Each toss is an independent event. So three heads in a row (heads AND heads AND heads) would have a probability of:1/2 * 1/2 * 1/2 = (1/2)^3 = 1/(2^3) = 1/8 = 12.5%
What is the probability of flipping three coins and all of them land up on heads?
The probability of the first coin landing heads is half (or 1/2). Similarly, the probability of the second and third coins landing heads are also 1/2 in each case. Therefore, the probability of having three heads is: (1/2)(1/2)(1/2) = (1/8)
Probability of heads in a coin toss?
1/2 Because there is one side with heads (1/) and the total sides (/2)
What is the probability of a coin tossed four times to get four heads?
The probability of a fair coin to land head is 1/2. Since for 4 flips it must land heads each time, the probability of 4 heads is 1/2 * 1/2 * 1/2 * 1/2 = 1/16.
What is the probability of getting tails on the first toss and heads on the second toss?
1 in 2 to get tails on first toss 1 in 2 to get heads on second toss To get tails on first toss and heads on second toss, probability is 1/2 x 1/2 = 1 in 4
A coin is tossed 9 times. What is the probability of getting exactly 4 heads.... What is the probability of getting at most 2 heads?
The number of trials: n = 9, the number of success: r = 4. The probability of success: P(H) = 1/2, the probability of failure: P(T) = 1/2. P(exactly 4 heads in 9 tosses) = 9C4 x (1/2)^5 x (1/2)^4 = 126(1/2)^9 = 0.246 P(at most 2 heads) = P(none heads) + P( 1 head) + P(2 heads) = 9C0[(1/2)^9][(1/2)^0] + 9C1[(1/2)^8][(1/2)^1] + 9C2[(1/2)^7][(1/2)^2] = 1(1/2)^9 + 9(1/2)^9 + 36(1/2)^9 = (1 + 9 + 36)(1/2)^9 = 46(1/2)^9 = 0.09
What is the probability of getting exactly 3 heads if a fair coin is tossed 8 times?
The probability to get heads once is 1/2 as the coin is fair The probability to get heads twice is 1/2x1/2 The probability to get heads three times is 1/2x1/2x1/2 The probability to get tails once is 1/2 The probability to get tails 5 times is (1/2)5 So the probability to get 3 heads when the coin is tossed 8 times is (1/2)3(1/2)5=(1/2)8 = 1/256 If you read carefully you'll understand that 3 heads and 5 tails has the same probability than any other outcome = 1/256 As the coin is fair, each side has the same probability to appear So the probability to get 3 heads and 5 tails is the same as getting for instance 8 heads or 8 tails or 1 tails and 7 heads, and so on
What is the probability of getting heads when you toss a coin?
50% or 1/2. There is 1 heads on a coin (numerator) There are 2 sides on a coin (denominator)
What is the probability that the first two flips will both be heads and the third flip will be either heads or tails if you flip three fair coins?
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
