14.Probability
easy

ત્રણ સિક્કા એક વાર ઉછાળવામાં આવે છે. નીચે આપેલ ઘટનાની સંભાવના શોધો.

વધુમાં વધુ  $2$ છાપ મળે. 

Option A
Option B
Option C
Option D

Solution

When three coins are tossed once, the sample space is given by $S =\{ HHH , HHT , HTH , THH , HTT , THT , TTH , TTT \}$

$\therefore$ Accordingly, $n ( S )=8$

It is known that the probability of an event $A$ is given by

$P ( A )=\frac{\text { Number of outcomes favourable to } A }{\text { Total number of possible outcomes }}=\frac{n( A )}{n( S )}$

Let $E$ be the event of the occurrence of at most $2$ heads.

Accordingly, $E =\{ HHT , \,HTH , \,THH , \,HTT , \,THT \,, TTH , \,TTT \}$

$\therefore P(E)=\frac{n(E)}{n(S)}=\frac{7}{8}$

Standard 11
Mathematics

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