**Question 1: **The number of tosses of a coin that are needed so that the probability of getting atleast one head be on 0.875 is

- 2
- 3
- 4
- 5

** Solution: **
Let the required number of tosses of the coin be n.

Then, P[Atleast one head in n
tosses of a coin] = 1 - P (Number of heads in n tosses of a coin)

= 1 - (

$\frac{1}{2}$)$^{n}$

We want n so that, this probability is 0.875

1 - (

$\frac{1}{2}$)$^{n}$ = 0.875

(

$\frac{1}{2}$)$^{n}$ = 1 - 0.875

= 0.125

= (0.5)$^{3}$

= (

$\frac{1}{2}$)$^{3}$

n = 3.

Therefore, option 2 is the correct solution.

**Question 2: **Assume that 0.5 of the population is vegetarian so that, the chance of an individual being a vegetarian is 0.5. Assuming that 100 investigators each take the sample of 10 individuals to see whether they are vegetarian, how many investigators would expect to report that three or less were vegetarian?

** Solution: **
n = 10

p = Probability that an individual is vegetarian is 0.5, q = 1 - p = 0.5

Then, by binomial probability law, the probability that there are 'r' vegetarian in a sample of 10 is

p(r) = $\binom{10}{r}p^{r}q^{10-r}$

= $\frac{1}{1024}\binom{10}{r}$

Thus, the probability that in a sample of 10, three or less people are vegetarian is

P(0) + P(1) + P(2) + P(3)

= $\frac{1}{1024}$ ($\binom{10}{0}\binom{10}{1}\binom{10}{2}\binom{10}{3}$)

= $\frac{1}{1024}$(1 + 10 + 45+ 120)

= $\frac{176}{1024}$

= $\frac{11}{64}$

Hence, out of 100 investigators, the number of investigators who will report three or less vegetarians in a sample of 10 is

= 100 x $\frac{11}{64}$

= $\frac{275}{16}$

= 17.2 $\simeq $ 17 (Since the number of investigators cannot be a fraction.)