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Nine cards (identical in all respects) are numbered \[\mathbf{2}\text{ }\mathbf{to}\text{ }\mathbf{10}\]. A card is selected from them at random. Find the probability that the card selected will be: \[\left( \mathbf{i} \right)\]an even number \[\left( \mathbf{ii} \right)\] a multiple of \[\mathbf{3}\]
Nine cards (identical in all respects) are numbered \[\mathbf{2}\text{ }\mathbf{to}\text{ }\mathbf{10}\]. A card is selected from them at random. Find the probability that the card selected will be: \[\left( \mathbf{i} \right)\]an even number \[\left( \mathbf{ii} \right)\] a multiple of \[\mathbf{3}\]
CBSE | Class 10 | Exercise 13.2 | Selina | Statistics and Probability
Nine cards (identical in all respects) are numbered \[\mathbf{2}\text{ }\mathbf{to}\text{ }\mathbf{10}\]. A card is selected from them at random. Find the probability that the card selected will be: \[\left( \mathbf{i} \right)\]an even number \[\left( \mathbf{ii} \right)\] a multiple of \[\mathbf{3}\]

Solution:

We know that, there are totally \[9\] cards from which one card is drawn.

Total number of elementary events \[=\text{ }n\left( S \right)\text{ }=\text{ }9\]

\[\left( i \right)\] From numbers \[2\text{ }to\text{ }10\], there are \[5\] even numbers i.e. \[2,\text{ }4,\text{ }6,\text{ }8,\text{ }10\]

So, favorable number of events \[=\text{ }n\left( E \right)\text{ }=\text{ }5\]

Hence, probability of selecting a card with an even number \[=\text{ }n\left( E \right)/\text{ }n\left( S \right)\text{ }=\text{ }5/9\]

\[\left( ii \right)\] From numbers \[2\text{ }to\text{ }10\], there are \[3\] numbers which are multiples of  i.e\[.\text{ }3,\text{ }6,\text{ }9\]

So, favorable number of events \[=\text{ }n\left( E \right)\text{ }=\text{ }3\]

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