A fair coin is tossed 5 times and comes up heads four times out of five. If the coin is tossed a sixth time under the same conditions, the probability of it turning up heads is
Answers
You can read “at least” also as probability = 1 - p(H=0) - p(H=1). In this case this calculation is simpler than p(H=2) + p(H=3) + p(H=4) + p(H=5), but they both yield the same result.
In order to calculate the result for this simple example we actually can just count all 32 results and see what fits.
HHHHH;
THHHH, HTHHH, HHTHH, HHHTH, HHHHT;
TTHHH, THTHH, THHTH, THHHT, HTTHH, HTHTH, HTHHT, HHTTH, HHTHT, HHHTT;
TTTHH, TTHTH, TTHHT, THTTH, THHTT, THTHT, HTTTH, HTTHT, HTHTT, HHTTT;
TTTTH, TTTHT, TTHTT, THTTT, HTTTT;
TTTT;
I felt it’s important to do this, because none of the previous answers had this result. So after counting up you see that the answer is 26/32 = 81.25%.
But it’s also important to understand the calculation behind it. In this case we call the calculation to be carried out “bigger number: total flips” Choose (C) “shorter number: flips we are interested in”, so “Total” Choose “Selection”.
In this case we have 5 flips total, and as we look at p(H=0) we see that there is only 1 combination 5C0 with zero “H” (5 Choose 0) and for p(H=1) there are 5 different combinations for 1 “H” 5C1 (5 Choose 1).
cf. your calculator for this function “C”. What does it do? It selects all the possible combinations to position them, which is 1x2x3x4x5 = 5! and then looks at combinations that have the same number of “H”, but a different order.
Possible results of 5C0 = 5! / 0! x (5–0)! = 5! / 1 x 5! = 1
Possible results of 5C1 = 5! / 1! x (5–1)! = 5! / 1 x 4! = 5
Possible results of 5C2 = 5! / 2! x (5–2)! = 5! / (2 x 3!) = 4 x 5 / 2 = 10
Due to axial symmetry the results of 5C3 = 10, 5C4 = 5, 5C5 = 1 result accordingly.
If you now calculate the probabilities, we have either 1 - 1/32 -5/32 = 26/32 or also the longer route for 10/32 + 10/32 + 5/32 + 1/32 =26/32. We get the same result as above with p = 81.25%
Answer:
Step-by-step explanation:
Since it’s a fair coin, both heads (H) and tails (T) have probability p = 0.5. After tossing it five times, we have 2^5 = 32 different possible outcomes. This also considers the order in which they appear (which doesn’t matter to us).
You can read “at least” also as probability = 1 - p(H=0) - p(H=1). In this case this calculation is simpler than p(H=2) + p(H=3) + p(H=4) + p(H=5), but they both yield the same result.
In order to calculate the result for this simple example we actually can just count all 32 results and see what fits.
HHHHH;
THHHH, HTHHH, HHTHH, HHHTH, HHHHT;
TTHHH, THTHH, THHTH, THHHT, HTTHH, HTHTH, HTHHT, HHTTH, HHTHT, HHHTT;
TTTHH, TTHTH, TTHHT, THTTH, THHTT, THTHT, HTTTH, HTTHT, HTHTT, HHTTT;
TTTTH, TTTHT, TTHTT, THTTT, HTTTT;
TTTT;