Question

Find the probability of getting 2 or 3 tails when a coin is tossed four times.

Answer 1

When a coin is tossed 4 times. A total of 2⁴ = 16 outcomes are possible. 

Let S be the set consisting of all such outcomes. 

∴ n(S) = 16 

Let A be the event of getting 2 tails. 

∴ A = {TTHH,THTH,THHT,HTTH,HTHT,HHTT} 

∴ n(A) = 6 

∴ P(A) = \(\frac{6}{16} \)= \(\frac{3}{8}\) 

Let B be the event of getting 3 tails. 

∴ B = { TTTH ,TTHT, THTT,HTTT } 

⇒ n(B) = 4

∴ P(B) = \(\frac{4}{16}\)= \(\frac{1}{4}\)

We need to find the probability of getting 2 tails or 3 tails i.e. 

P(A∪B) = ? 

As we can’t get 2 and 3 tails at the same time. So A and B are mutually exclusive events. 

∴ P(A∪B) = P(A) + P(B) =  \(\frac{3}{8}\) + \(\frac{1}{4}\)= \(\frac{5}{8}\)