( A fair coin is tossed 10 times, find the
probability of obtaining (i) exactly six heads
(ii) at least six heads (iii) at most six heads.
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a) Probability of exactly six heads=10
C
6
(0.5)
10
b)Probability of at least 6 heads=
P(6 heads)+P(7 heads)+P(8 heads)+P(9 heads)+P(10 heads)
=10
C
6
(0.5)
10
+10
C
7
(0.5)
10
+10
C
8
(0.5)
10
+10
C
9
(0.5)
10
+10
C
10
(0.5)
10
=386/1024
c))Probability of at most 6 heads=
1-P(7 heads)+P(8 heads)+P(9 heads)+P(10 heads)
=1−10
C
7
(0.5)
10
+10
C
8
(0.5)
10
+10
C
9
(0.5)
10
+10
C
10
(0.5)
10
=848/1024
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