A fair coin is tossed 4 times. what is the probability that at least 2 heads appear
1 answer:
HHHH TTTT HHHT HHTH HTHH THHH TTTH TTHT THTT HTTT
HHTT HTHT HTTH TTHH THTH THHT
- is the sample space.
Total possible arrangements = 16.
Number with at least 2 heads = 11
Required probability = 11/16
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Answer:
now it would be $78
(-12)1 because of how you break it down
0.941176471 to the nearest hundredth is 0.94
80n = 68*100
80n = 6800
n = 6800/80
n = 85
6y-4=4y+2
2y=6, so y=3.
Plugging it in, you get that AC equals 4*3+2=14.
This is an isosceles triangle.
