Three unbiased coins are tossed. What is the probability of getting at most two heads?
A. $$\frac{{3}}{{4}}$$
B. $$\frac{{1}}{{4}}$$
C. $$\frac{{3}}{{8}}$$
D. $$\frac{{7}}{{8}}$$
Answer: Option D
Solution (By Examveda Team)
Getting at most Two heads means 0 to 2 but not more than 2Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at most two heads
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}
$$\therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{7}{8}$$
This Question Belongs to Arithmetic Ability >> Probability
The answer is 3/4 not 7/8 because TTT is not in n(E)
At most two means 0,1,2 but not more than Two. so (TTT,TTH,THT,HTT,THH,HTH,HHT=7) But no HHH not acceptable because here 3 head. Probability=7/8
Answer is 3/4.Because n(E)= 6. so .6÷8 = 3/4.ok sir.
I think ans 3/8
Here said that probability of getting at most two heads?
N(E)={HHH},{HHT},{HTH},{THH}
so probability is= 4/8 or 1/2
I think 3/8
how 7/8 is the answer