Solution 6:
a)
b)
Solution 7:
A balanced coin is tossed four times, So the possible outcomes can be following:
possible outcomes are 2 and trials are 4
Sample space is determined by= = 16
HHHH HTHH THHH HTHT
HHHT HTTH TTHH THTH
HHTT HHTH TTTH THHT
HTTT TTTT TTHT THTT
Probability = Number of events / Total numbers of events
A = Atleast 3 heads
A = {HHHH HTHH THHH HHHT HHTH}
B = Atmost two heads
B = {HTHT HTTH TTHH THTH HHTT TTTH THHT HTTT TTTT TTHT THTT}
C = heads on the third toss
C = {HHHH HTHH THHH HTHT HHHT TTHH THHT TTHT}
D = 1 head and 3 tails
D = {TTTH HTTT TTHT THTT}
a)
A = {HHHH HTHH THHH HHHT HHTH}
P(A) = 5/16
0.3125
b)
Null
c)
B = {HTHT HTTH TTHH THTH HHTT TTTH THHT HTTT TTTT TTHT THTT}
P(B) = 11/16
0.6875
d)
{HHHH HTHH THHH HHHT}
0.25
e)
D = {TTTH HTTT TTHT THTT}
P(D) = 4/16
0.25
f)
= {HHHH HTHH THHH HTHT HHHT TTHH THHT TTHT HHTH}
0.5625
g)
= {TTTH HTTT TTHT THTT}
0.25
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