multiple Nine cards (identical in all respects) are numbered . A card is selected from them at random. Find the probability that the card selected will be:

    \[\left( \mathbf{iii} \right)\]

an even number and a multiple of

    \[\mathbf{3}\]

    \[\left( \mathbf{iv} \right)\]

an even number or a of

    \[\mathbf{3}\]

multiple Nine cards (identical in all respects) are numbered . A card is selected from them at random. Find the probability that the card selected will be:

    \[\left( \mathbf{iii} \right)\]

an even number and a multiple of

    \[\mathbf{3}\]

    \[\left( \mathbf{iv} \right)\]

an even number or a of

    \[\mathbf{3}\]

Solution:

    \[\left( iii \right)\]

From numbers

    \[2\text{ }to\text{ }10\]

, there is one number which is an even number as well as multiple of

    \[3\text{ }i.e.\text{ }6\]

So, favorable number of events

    \[=\text{ }n\left( E \right)\text{ }=\text{ }1\]

Hence, probability of selecting a card with a number which is an even number as well as multiple of

    \[3\]

    \[=\text{ }n\left( E \right)/\text{ }n\left( S \right)\text{ }=\text{ }1/9\]

    \[\left( iv \right)\]

From numbers

    \[2\text{ }to\text{ }10\]

, there are

    \[7\]

numbers which are even numbers or a multiple of

    \[3\]

i.e.

    \[2,\text{ }3,\text{ }4,\text{ }6,\text{ }8,\text{ }9,\text{ }10\]

So, favorable number of events

    \[=\text{ }n\left( E \right)\text{ }=\text{ }7\]

Hence, probability of selecting a card with a number which is an even number or a multiple of

    \[3\]

    \[=~n\left( E \right)/\text{ }n\left( S \right)\text{ }=\text{ }7/9\]