Question

When tossing 4 coins simultaneously,find the probability at least 1 head is showing?

Answer 1

there are 4 coins.
Each coin has a \[\frac{1} { 2 }\]chance.
you got it so far?

Answer 2

so it would be 1/2*4 =2 ??

Answer 3

kind of. just look.

Answer 4

coin 1
\[\frac{1} {2}\]
coin 2
\[\frac{1} {2}\]
coin 3
\[\frac{1} {2}\]
coin 4
\[\frac{1} {2}\]

Answer 5

wouldn't be 50% percent ?

Answer 6

that is the chance of getting heads on all for of them.
so then it would be:
\[\frac{4} {2}\]
so yea, you were right, my bad. cuz' when you simplify it would become:
\[\frac{2} {1}\]

Answer 7

when 4 coins are tossed only one event does not have head which is TTTT.
T=tails
so the probability will be 15/16

Answer 8

no its 15/16 ..
binomial ..
u get 1 head by this no. of wasys = 4C1\[\left( \frac{ 1 }{ 2 } \right)^{4}\]
u get 2 head by this no. of wasys = 4C2\[\left( \frac{ 1 }{ 2 } \right)^{4}\]
u get 3 head by this no. of wasys = 4C3\[\left( \frac{ 1 }{ 2 } \right)^{4}\]
u get 4 head by this no. of wasys = 4C4\[\left( \frac{ 1 }{ 2 } \right)^{4}\]

total no. of ways = (adding the above quantities)
= 15/16

Answer 9

its 4C1 * (1/2)^4

Answer 10

okay, im in algebra 3, i just finished probablility, how dahell did you get \[\frac{15}{16}\]

Answer 11

i m in 10+2 ... man this is the basic problem ...
probaly the first question of the topic..

Answer 12

google binomial distribution and then u can study it ..
may be helpful..