Question

A coin is tossed 80 times with the following outcomes:
Head: 35 times and Tail: 45 times.
In a single throw of a coin, if E be the event of getting a head, verify that P(E) + P(not E) = 1.

Answer 1

Total number of trials = 80
Number of times the tail appears = 45

So, number of times the head appears = 35

Let E be the event of getting a head. Then,
P(getting a head) = P (E) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}\mathrm{head}\mathrm{appears}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{35}{80}=\frac{7}{16}$

P(not E) = P (getting a tail) = ​$\frac{\mathrm{Number}\mathrm{of}\mathrm{times}\mathrm{tail}\mathrm{appears}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{45}{80}=\frac{9}{16}$

P (E) + ​P(not E) = $\frac{7}{16}+\frac{9}{16}=\frac{16}{16}=1$
Hence, for an event E of getting a head, P(E) + P(not E) = 1