1 out of 555555
If the order of the outcomes matters, then TTTT, TTTH, TTHT, THTT, HTTT, TTHH, THTH, THHT, HTTH, HTHT, HHTT, THHH, HTHH, HHTH, HHHT, HHHH. If the order does not matter, then TTTT, TTTH, TTHH, THHH AND HHHH
There are 16 possible outcomes, each of which is equally likely. So each has a probability of 1/16 HHHH: X = H*T = 4*0 = 0 HHHT: X = H*T = 3*1 = 3 HHTH: X = H*T = 3*1 = 3 HTHH: X = H*T = 3*1 = 3 THHH: X = H*T = 3*1 = 3 HHTT: X = H*T = 2*2 = 4 HTHT: X = H*T = 2*2 = 4 HTTH: X = H*T = 2*2 = 4 THHT: X = H*T = 2*2 = 4 THTH: X = H*T = 2*2 = 4 TTHH: X = H*T = 2*2 = 4 HTTT: X = H*T = 1*3 = 3 THTT: X = H*T = 1*3 = 3 TTHT: X = H*T = 1*3 = 3 TTTH: X = H*T = 1*3 = 3 TTTT: X = H*T = 0*4 = 0 So, the probability distribution function of X is f(X = 0) = 1/16 f(X = 1) = 4/16 = 1/4 f(X = 2) = 6/16 = 3/8 f(X = 3) = 4/16 = 1/4 f(X = 4 = 1/16 and f(X = x) = 0 for all other x
The term 'sample space' can be somewhat arbitrary. In this case, it might be any of the following (or another) possibility: If order of flips is significant: HHHH HHHT HHTH HHTT HTHH HTHT HTTH HTTT THHH THHT THTH THTT TTHH TTHT TTTH TTTT If order is not significant: HHHH HHHT HHTT HTTT TTTT If, say, only the number of either heads or tails is important: 4 3 2 1
Oh, what a happy little question! Let's break it down. The probability of getting heads on a single flip is 1/2, and the probability of getting tails is also 1/2. So, the probability of getting HTTH in that specific order on four flips would be (1/2) * (1/2) * (1/2) * (1/2) = 1/16. Just remember, there are no mistakes, just happy little accidents in probability!
There are 16 permutations of a coin tossed 4 times. They are ...HHHH - HHHT - HHTH - HHTTHTHH - HTHT - HTTH - HTTTTHHH - THHT - THTH - THTTTTHH - TTHT - TTTH - TTTT... and the permutations with only one tail are bolded. There are 4, so the probability of flipping a coin 4 times and getting only 1 tail is 4 in 16, or 1 in 4, or 0.25.The coin does not have a memory therefore the probability is always 50/50.
HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT
HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT.
If the order of the outcomes matters, then TTTT, TTTH, TTHT, THTT, HTTT, TTHH, THTH, THHT, HTTH, HTHT, HHTT, THHH, HTHH, HHTH, HHHT, HHHH. If the order does not matter, then TTTT, TTTH, TTHH, THHH AND HHHH
There are 16 possible outcomes, each of which is equally likely. So each has a probability of 1/16 HHHH: X = H*T = 4*0 = 0 HHHT: X = H*T = 3*1 = 3 HHTH: X = H*T = 3*1 = 3 HTHH: X = H*T = 3*1 = 3 THHH: X = H*T = 3*1 = 3 HHTT: X = H*T = 2*2 = 4 HTHT: X = H*T = 2*2 = 4 HTTH: X = H*T = 2*2 = 4 THHT: X = H*T = 2*2 = 4 THTH: X = H*T = 2*2 = 4 TTHH: X = H*T = 2*2 = 4 HTTT: X = H*T = 1*3 = 3 THTT: X = H*T = 1*3 = 3 TTHT: X = H*T = 1*3 = 3 TTTH: X = H*T = 1*3 = 3 TTTT: X = H*T = 0*4 = 0 So, the probability distribution function of X is f(X = 0) = 1/16 f(X = 1) = 4/16 = 1/4 f(X = 2) = 6/16 = 3/8 f(X = 3) = 4/16 = 1/4 f(X = 4 = 1/16 and f(X = x) = 0 for all other x
The term 'sample space' can be somewhat arbitrary. In this case, it might be any of the following (or another) possibility: If order of flips is significant: HHHH HHHT HHTH HHTT HTHH HTHT HTTH HTTT THHH THHT THTH THTT TTHH TTHT TTTH TTTT If order is not significant: HHHH HHHT HHTT HTTT TTTT If, say, only the number of either heads or tails is important: 4 3 2 1