Question

In 256 sets of 12 tosses of a coin, in how many cases one can expect 8 heads and 4 tails?

Answer 1

Final Answer : 31

Steps:
1) We will use Binomial Distribution of events,

Probability of getting 8 heads and 4 tails in 12 tosses in 1 trial round

P (8 Heads, 4 tails in 12 tosses)

= Selecting 8 heads out of 12 tosses * Probability of getting head in all 8 tosses * Selecting 4 tails from remaining tosses * Probability of getting tail in all remaining tosses
$\binom{12}{8} \times \frac{1}{ {2}^{8} } \times \binom{4}{4} \times \frac{1}{ {2}^{4} } \\ = > \binom{12}{8} \times \frac{1}{ {2}^{12} }$

2) But, we have 256 trial rounds,
so No. of cases in which we can expect 8 heads and 4 tails is given by :

P (8 Heads, 4 Tails in 12 Tosses) * 256

$= > \binom{12}{8} \times \frac{1}{ {2}^{12} } \times 256$


This will be no. of cases .
I am lazy to do this Calculation, so used Scientific Calculator to do this stuff.

=> 30. 9375
=> 31 (approx.)

Hence, 31 cases are there in which we can expect our desired event.