Question

A fair coin is tossed $12$ times. Find the probability that two heads do not occur consecutively.

• A. Required probability is $\frac{375}{4096}$
• B. Required probability is $\frac{376}{4096}$
• C. Required probability is $\frac{377}{4096}$
• D. Required probability is $\frac{383}{4096}$

Answer 1

Answer: (C)

Required probability is $\frac{377}{4096}$

Total Possible outcomes = $2^{12}$
.No two heads together in $12$ throws,
so max. no. of possible heads = $6$

No. of heads	Possible combinations	Total
0	1	1
1	${}^{12}{C}_1$	12
2	${}^{11}{C}_2$	55
3	${}^{10}{C}_3$	120
4	${}^9{C}_2$	126
5	${}^8{C}_5$	26
6	${}^7{C}_6$	7
	Total	377

So
$P=\frac{377}{4096}$