Question

 A fair coin is tossed repeatedly. The probability of getting a result in the sixth toss different from those obtained in the first five tosses is
1. \(\frac{5}{16}\)
2. \(\frac{1}{12}\)
3. \(\frac{1}{64}\)
4. \(\frac{5}{64}\)
5. \(\frac{1}{32}\)

Answer 1

Correct Answer - Option 5 : \(\frac{1}{32}\)

Concept:

\(\rm P(E)= \frac{n(E)}{n(S)}\), where n(E) = No. of favourable cases for event A and n(S) = cardinality of sample space.

A coin tossed: P(getting head) = \(\frac12\), P(getting tail) = \(\frac12\)

P(A and B) = P(A) × P(B)

P(A or B) = P(A) + P(B)

 

Calculation:

A fair coin is tossed repeatedly.

Sixth toss different from those obtained in the first five tosses.

∴ The possibility is that in the first five tosses, we get 5 heads and 1 tail OR five tails and one head 

 

Let, Hi denotes event getting 'head' in 'i' th toss and Ti denotes event getting 'tail' in 'i' th toss.

So, required probability = [\(\rm P(H_1∩ H_2∩ H_3∩ H_4∩ H_5∩ T_6)\)OR \(\rm P(T_1∩ T_2∩ T_3∩ T_4∩ T_5∩ H_6)\)]

= {\(\rm [P(H_1∩ H_2∩ H_3∩ H_4∩ H_5∩ T_6)]\) ∪ \(\rm [P(T_1∩ T_2∩ T_3∩ T_4∩ T_5∩ H_6)]\)} 

= \(\rm [P(H_1)\times P( H_2)\times P(H_3)\times P(H_4)\times P( H_5) \times P(T_6)]\) + \(\rm [P(T_1)\times P(T_2) \times P(T_3) \times P( T_4) \times P( T_5) \times P( H_6)]\)

= \([\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}]\) + \([\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}]\) 

= \(\frac{1}{64}\) + \(\frac{1}{64}\)

= \(\frac{2}{64}\)

= \(\frac{1}{32}\)

Hence, option (2) is correct.