Question

Four coins are tossed simultaneously 810 times with the following frequencies of different outcomes.

No. of heads	4	3	2	1	0
Frequency	150	182	210	129	141

The probability of getting at least 2 tails is

• A. $\frac{16}{27}$
• B. $\frac{11}{18}$
• C. $\frac{2}{3}$
• D. $\frac{16}{37}$

Answer 1

The correct option is A $\frac{16}{27}$

Total number of times, four coins are tossed = 810
Outcomes having at least 2 tails: 2 tails, 3 tails or 4 tails i.e., 2 heads, 1 head or no head.
$\therefore \text{Number of outcomes getting at least 2 tails}=210+129+141$
= 480
$\therefore \text{Required Probability}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$
$=\frac{480}{810}$
$=\frac{16}{27}$
Hence, the correct answer is option (a).