We flip a fair coin 10 times. What is the probability that we get heads in at least 5 of the 10 flips?

Guest May 23, 2021

#1**+2 **

Total number of coin flip configurations: \(2^{10} = 1024\)

(2 choices for each flip: heads or tails, for 10 flips)

heads in 5 flips: \(10 \choose 5\)

(There are 10 choose 5 ways to rearrange the flips in the sequence HHHHHTTTTT.)

heads in 6 flips: \(10 \choose 6\)

heads in 7 flips: \(10 \choose 7\)

etc.

probability = \(\frac{{10 \choose 5} + {10 \choose 6} + {10 \choose 7} + {10 \choose 8} + {10 \choose 9} + {10 \choose 10}}{2^{10}} = \frac{628}{1024} = \boxed{\frac{157}{256}.} \)

CubeyThePenguin May 23, 2021

#2**0 **

Now, I am not sure of this answer, cus I'm kinda rusty in probability. In what I think, we start:

lif there are a half chance of heads or tails, then we put in TWO different things into the expression x+y. Since it's ten flips, we put it to the tenth power ((x+y)^10). Using Pascal's triangle, we get x^10+10x^9*y+45x^8*y^2+120x^7*y^3+210x^6*y^4+252x^5*y^5+210x^4*y^6+120x^3+y^7+45x^2*y^8+10xy^9+y^10 (sry I don't know latex try to enterpret it urself sry sry). In this case, x=heads (1/2) and y=tails (1/2). Since it's five heads or more heads, it is equal to if heads is five (the number where it's x^5), then x^6 and x^7 and x^8 and x^9 and x^10. These parts of it are: x^10+10x^9*y+45x^8*y^2+120x^7*y^3+210x^6*y^4+252x^5*y^5, but plugging in 1^2 for x and y, we get x=y. Therefore, since they are equal, we can combine them to one x, and since they all add up to 10, we just get x^10+10x^10+45x^10+120x^10+210x^10+252x^10. After, we combine like terms and get (1+10+45+120+210+252)x^10=638x^10. Finally, we plug in 1/2 for x and we get 638/1024. Then, we need to simplify this. We divide by two (since 1024's only factor is 2, and 638 only has 2^1 in it, we only divide by 2^1=2) on both the numerator and denominator. Then, we get 319/512. Hope this helped :) (although it probably didnt cus I don't know latex lol).

MathPerson.exe May 23, 2021