## Johnmathstudent 2 years ago If I flip four coins what is the probability that at least two of the coins are heads?

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1. Johnmathstudent

2. amistre64

its the opposite of P(at most 1 is heads)

3. amistre64

1 - P(0) - P(1)

4. MAYSTEPH

do you have choices

5. amistre64

there are 16 outcomes tttt ttth ttht thtt httt ... 1 4 6 4 1 = 16 0 heads is 1 out of 16 1 head is 4 out of 16

6. Johnmathstudent

7. amistre64

at least 2 heads means that you have 2, 3, or 4 heads P(0) + P(1) + P(2) + P(3) + P(4) = 1 P(2) + P(3) + P(4) = 1 - P(0) - P(1)

8. amistre64

since P(0) = 1/16 and P(1) = 4/16 the solution is determined

9. Johnmathstudent

So what would the answer be written as?

10. amistre64

im not going to do the subtraction for you ....

11. amistre64

you can tell me what you get ... and i can verify if you did it correctly

12. MAYSTEPH

bro your even confusing me.... amistre64

13. amistre64

im not a mind reader, so youll have to be more detailed in what is confusing :)

14. MAYSTEPH

nevermind .. the probability would be 50% John

15. amistre64

the probability is not 50%; for starters, a probability is a number between 0 and 1 ... not a percentage.

16. amistre64

and for another, 1 - P(0) - P(1) does not equal .5000 or 1/2

17. MAYSTEPH

he asked for a probability. if you flip 4 coins and the odds of 2 of the comming out heads is 2/4 and tht equals 50% and wtf is p

18. MAYSTEPH

its not rocket science he just wants a simple answer.. idk whats wrong with tht

19. amistre64

\[\binom{4}{0}.5^0.5^4+\binom{4}{1}.5^1.5^3+\binom{4}{2}.5^2.5^2+\binom{4}{3}.5^3.5^1+\binom{4}{4}.5^n.4^0=1\] \[(1).5^0.5^4+(4).5^1.5^3+(6).5^2.5^2+(4).5^3.5^1+(1).5^4.5^0=1\]

20. amistre64

P(E) is notation for "probability of an event"

21. amistre64

the last 3 terms are 4 choose 2, 4 choose 3, and 4 choose 4 which is what we want if we are to determine "at least 2" out of 4

22. amistre64

subtracting the first 2 terms from each side gives us the solution

23. MAYSTEPH

your making this way more complex than it already isnt lol if you know the damn answer then tell him and quit arguing with a degenerate!

24. amistre64

i suggest you tone down your language. and, we are not providing a free answering service, we are offering a way to learn the material.

25. amistre64

my first attempt was to simplify it .... since that did not work i resorted to the longer, more detailed version of it all.

26. Johnmathstudent

27. Johnmathstudent

Or 11/16

28. amistre64

11/16 is correct either way is fine, but the format is up to whoever is doing the grading.

29. Johnmathstudent

Thanks I just forgot how to do this from the beginning of the year

30. amistre64

you did fine :) good luck