Question

Three coins are tossed simultaneously 100 times with the following frequencies of different outcomes :

$\begin{array}{ccccc}\text{Outcome}& \text{No head}& \text{One head}& \text{Two heads}& \text{Three heads}\\ \text{Frequency}& 14& 38& 36& 12\end{array}$

If the three coins are simultaneously tossed again, compute the compute the probability of :

(i) 2 heads coming up. (ii)3 heads coming up.

(iii) at least one head coming up. (iv) getting more heads than tails.

(v) getting more tails heads.

Answer 1

Three coins are tossed simultaneously 100 times

$\therefore$ Total out comes (n) = 100

$\begin{array}{ccccc}\text{Outcome}& \text{No head}& \text{One head}& \text{Two heads}& \text{Three heads}\\ \text{Frequency}& 14& 38& 36& 12\end{array}$

(i) Probability of 2 heads coming up (m) =36

$\therefore$ P(A) =$\frac{m}{n}=\frac{36}{100}=0.36$

(ii) probability of 3 heads (m) =12

$\therefore$ P(A) =$\frac{m}{n}=\frac{12}{100}=0.12$

(iii) Probability of at least one heads coming up (m) =38+36+12=86

$\therefore$ P(A) =$\frac{m}{n}=\frac{86}{100}=0.86$

(iv) Probability of getting more heads than tails (m) =36+12=48

$\therefore$ P(A) =$\frac{m}{n}=\frac{48}{100}=0.48$

(v) Getting more tails than heads (m) =14+38=52

$\therefore$ P(A) =$\frac{m}{n}=\frac{52}{100}=0.52$