You can flip a coin twice find the probability that at least one of the tosses showed heads

3/4

Explanation:

First I will write sample space for 2 time tossing :

S={(T,T),(H,T),(H,H),(T,H)} ...where(T,T)means you get tail in first attempt and in second time you get tail...and similarly understand (H,T),(H,H) by same logic.

So, probability that atleast one toss shows head is,

P(H>=1) why,because the least you can get is 1 head and the condition doesn't limits you to get more than 1 head.

So,

P(H>=1)=P(H=1)+P(H=2)    here H more than 2 not possible

            =2/4 +1/4

            =3/4

P(H=1) is equal to 2/4 because in sample space there are two events where head is obtained 1 time which is (T,H) and (H,T).Thus 2 divided by total events (i.e 4) =2/4

P(H=2) follows same logic

the answer is c hope it

step-by-step explanation: