+0  
 
0
102
4
avatar

We flip a fair coin 10 times. What is the probability that we get heads in at least 6 of the 10 flips?

 Mar 20, 2020
 #1
avatar+4569 
0

Each coin has two sides, heads or tails, so the total number of possible outcomes is \(2^{10}=1024\).

 

The number of ways of getting heads is 10 C 6   *   10  C  7   + 10   C  8  +  10 C  9   + 10  C  10=210+120+45+10+1=386 ways.

 

 

193/512  but I might be worng

 Mar 20, 2020
 #2
avatar
0

10! / [6!*4!*2^10] + 10! / [7!*3!*2^10] + 10! / [8!*2!*2^10] + 10! / [9!*1!*2^10] + 10! / [10!*2^10] =193 / 512 =37.6953125%

 Mar 20, 2020
 #3
avatar+111360 
+2

At least 6  heads means

 

6 heads  =  C(10,6) (,5)^10

7 heads = C(10, 7) (.5)^10

8 heads   = C(10,8)(.5)^10

9 heads  = C(10,9) (,5)^10

10 heads = (.5)^10

 

So....the total probability is

 

(.5)^10  [  C(10,6) +C(10,7) + C(10,8) + C(10,9)  + 1  ]     ≈ .3769  ≈  37.69%

 

 

cool cool cool

 Mar 20, 2020
 #4
avatar+1116 
+1

6H, 4T = 10!/6!4! ways to arrange

7H, 3T = 10!/7!3!

8H, 2T = 10!/8!/2!

9H, 1T = 10!/9!1!

10H, 0T = 10!/10!0!

Add it up to get 386

2^10 possible outcomes = 1024

Answer: 193/512

 

You are very welcome!

:P

 Mar 20, 2020

22 Online Users

avatar
avatar
avatar
avatar