The probability of getting atmost two heads when three coins are tossed is
A) 1/8
B) 8
C) 7/8
D) 3/8
Call
Correct option is (C) 7/8
If three coins are tossed then possible outcomes are S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
\(\therefore\) Total outcomes is n(S) = 8
Let event E be event of getting at most two heads.
\(\therefore\) E = {HHT, HTH, THH, HTT, THT, TTH, TTT}
\(\therefore\) n(E) = 7
\(\therefore\) P(E) \(=\frac{n(E)}{n(S)}=\frac78\)
Hence, the probability of getting at most two heads is \(\frac{7}{8}.\)
Call