Question

three coins are tossed simultaneously, find the probability of getting atleast two hands ​

Answer 1

Answer:

1/2

Step-by-step explanation:

sample space of tossing three coins

simultaneously =>

{ HHH , HHT , HTH , HTT , THH , THT , TTH , TTT }

Getting atleast two heads = { HHH , HHT , HTH , THH }

Probability of getting atleast two heads =

Number of favourable outcomes / Total number of outcomes

= 4 / 8

= 1 / 2

hope it helps you

Answer 2

$\huge \mathfrak \orange {\underline{Answer}}$

Sample Space = {HHH, HHT, HTT, HTH, TTT, TTH, THH, THT}

$\Large \text{Probability (E)=} \frac{Favourable Outcomes}{Total Outcomes}$

Probability of getting atleast 2 heads= 4

Total outcomes= 8

$\Large \text \pink{Substituting\: the\: values}$

$\text{Probability (E)=} \frac{4}{8}$

$\text{Probability (E)=} \frac{1}{2}$

Hope it helps úh

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