Question

A fair coin is tossed three times and the events A, B, and C are defined as follows: A:{ At least one head is observed } B:{ At least two heads are observed } C:{ The number of heads observed is odd } Find the following probabilities by summing the probabilities of the appropriate sample points (note that 0 is an even number):
(a) P(B) =
(b) P(A or B) =
(c) P(A or B or C)

Answer 1

Answer:

(a) 1/2

(b) 1/2

(c) 1/8

Step-by-step explanation:

Since, when a fair coin is tossed three times,

The the total number of possible outcomes

n(S) = 2 × 2 × 2

= 8 { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT },

Here, B : { At least two heads are observed } ,

⇒ B = {HHH, HHT, HTH, THH},

⇒ n(B) = 4,

Since,

$\text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}$

(a) So, the probability of B,

$P(B) =\frac{n(B)}{n(S)}=\frac{4}{8}=\frac{1}{2}$

(b) A : { At least one head is observed },

⇒ A = {HHH, HHT, HTH, THH, HTT, THT, TTH},

∵ A ∩ B = {HHH, HHT, HTH, THH},

n(A∩ B) = 4,

$\implies P(A\cap B) = \frac{n(A\cap B)}{n(S)} = \frac{4}{8}=\frac{1}{2}$

(c) C: { The number of heads observed is odd },

⇒ C = { HHH, HTT, THT, TTH},

∵ A ∩ B ∩ C = {HHH},

⇒ n(A ∩ B ∩ C) = 1,

$\implies P(A\cap B\cap C)=\frac{1}{8}$