(C) Total cards $= 30$. Number of white cards $= 17$. Number of green cards $= 30 - 17 = 13$. Number of $IMPORTANT$ cards $= 4 + 5 = 9$. Number of car

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(C) Total cards $= 30$. Number of white cards $= 17$. Number of green cards $= 30 - 17 = 13$. Number of $IMPORTANT$ cards $= 4 + 5 = 9$. Number of cards that are both green and $IMPORTANT = 5$. Let $G$ be the event of choosing a green card and $I$ be the event of choosing an $IMPORTANT$ card. We need to find $P(G \cup I) = P(G) + P(I) - P(G \cap I)$. $P(G) = \frac{13}{30}$,$P(I) = \frac{9}{30}$,$P(G \cap I) = \frac{5}{30}$. $P(G \cup I) = \frac{13}{30} + \frac{9}{30} - \frac{5}{30} = \frac{17}{30}$.

Class 11 Mathematics Probability Set Based probability

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In a set of $30$ game cards,$17$ are white and the rest are green. Out of $30$,$4$ white and $5$ green cards are marked $IMPORTANT$. If a card is chosen randomly from this set,then the probability of choosing a green card or an $IMPORTANT$ card is:

AP EAMCET 2022 All AP EAMCET

A
$13 / 30$
B
$22 / 30$
C
$17 / 30$
D
$9 / 13$

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In a set of $30$ game cards,$17$ are white and the rest are green. Out of $30$,$4$ white and $5$ green cards are marked $IMPORTANT$. If a card is chosen randomly from this set,then the probability of choosing a green card or an $IMPORTANT$ card is:

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