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Mathematics [2 Mark Question Answer]

Probability · ICSE STD 10 (ICSE - English Medium). 18 questions.

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[2 Mark Question Answer]

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18 questions · timed · auto-graded

Question 12 Marks
A card is drawn from a pack of well - shuffled 52 playing cards . Find the probability that the card drawn is
(vi) a card bearing a number between 2 and 6 including both.
Answer
There are 52 playing cards in a pack.
Number of cards between 2 and 6 induding both $=20$
$
\mathrm{P}(\text { drawing a card between } 2 \text { and } 6 \text { including both })=\frac{\text { Favourable outcomes }}{\text { Total outcomes }}=\frac{20}{52}=\frac{5}{13}
$
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Question 22 Marks
A card is drawn from a pack of well-shuffled 52 playing cards. Find the probability that the card is drawn is a face card.
Answer
There are 52 playing cards in a pack.
Number of face cards $=12$
$
\mathrm{P}(\text { drawing a face card })=\frac{\text { Favourable outcomes }}{\text { Total outcomes }}=\frac{12}{52}=\frac{3}{13}
$
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Question 32 Marks
A card is drawn from a pack of well - shuffled 52 playing cards . Find the probability that the card drawn is
(iv) an ace or a queen
Answer
There are 52 playing cards in a pack.
Number of ace cards $=4$ and number of queens $=4$
$
\mathrm{P}(\text { drawing an ace or a queen })=\frac{\text { Favourable outcomes }}{\text { Total outcomes }}=\frac{4}{52}+\frac{4}{52}=\frac{1}{13}+\frac{1}{13}=\frac{2}{13}
$
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Question 42 Marks
A card is drawn from a pack of well - shuffled 52 playing cards . Find the probability that the card drawn is
(iii) a king
Answer
There are 52 playing cards in a pack.
Number of kings $=4$
$
P(\text { drawing a king })=\frac{\text { Favourable outcomes }}{\text { Total outcomes }}=\frac{4}{52}=\frac{1}{13}
$
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Question 52 Marks
A card is drawn from a pack of well - shuffled 52 playing cards . Find the probability that the card drawn is
(ii) a red card
Answer
There are 52 playing cards in a pack.
Number of red cards $=26$
$
\mathrm{P}(\text { drawing a red card })=\frac{\text { Favourable outcomes }}{\text { Total outcomes }}=\frac{26}{52}=\frac{1}{2}
$
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Question 62 Marks
A card is drawn from a pack of well - shuffled 52 playing cards . Find the probability that the card drawn is
(i) a diamond
Answer
There are 52 playing cards in a pack.
There are 13 diamond cards.
$\mathrm{P}($ card drawn is a diamond $)=\frac{\text { favourable outcomes }}{\text { Total outcomes }}=\frac{13}{52}=\frac{1}{4}$
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Question 72 Marks
Two different coins are tossed simultaneously .Find the probability of getting :
(iii) atleast one tail
Answer
Possible outcomes for tossing two coins are: $\mathrm{HH}, \mathrm{HT}, \mathrm{TH}, \mathrm{TT}$ So the number of outcomes $=4$
favorable outcomes are: TT, TH, HT
$
P(\text { At least one tail })=\frac{3}{4}
$
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Question 82 Marks
Two different coins are tossed simultaneously .Find the probability of getting :
(ii)no tail
Answer
Possible outcomes for tossing two coins are: $\mathrm{HH}, \mathrm{HT}, \mathrm{TH}, \mathrm{TT}$ So the number of outcomes $=4$
favorable outcomes are: $\mathrm{HH}$
So, $P($ No tail $)=\frac{1}{4}$
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Question 92 Marks
Two different coins are tossed simultaneously .Find the probability of getting :
(i)two tails
Answer
Possible outcomes for tossing two coins are : $\mathrm{HH}, \mathrm{HT}, \mathrm{TH}, \mathrm{TT}$ So the number of outcomes $=4$
favorable outcomes are : TT
So, $P($ Two tails $)=\frac{1}{4}$
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Question 102 Marks
A box contains 5 red marbles, 8 white marbles and 4 grem marbles. One marble is taken out of the box at randam.
What is the probability that the marble taken out wil be
(ii) white?
Answer
Total number of marbles $=5+8+4=17$
Number of white marbles $=8$
$
\text { Probability of getting a white marble }=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{8}{17}
$
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Question 112 Marks
A box contains 5 red marbles, 8 white marbles and 4 grem marbles. One marble is taken out of the box at randam.
What is the probability that the marble taken out wil be
(i) red?
Answer
Total number of marbles $=5+8+4=17$
Number of red marbles $=5$
Probability of getting a red marble $=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{5}{17}$
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Question 122 Marks
f the probability of winning a game is 5/11,what is the probability of losing?
Answer
Since $\mathrm{P}($ Winning $)+\mathrm{P}($ Losing $)=1$
Therefore, $\mathrm{P}($ Losing $)=1-\mathrm{P}($ winning $)=1-\frac{5}{11}=\frac{6}{11}$
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Question 132 Marks
A bag contains a red ball , a blue ball and a yellow ball , all the balls being of the same size . Anjali takes out a ball from the bag without looking into it . what is the probability that she takes out .
(iii) blue ball?
Answer
Total Number of outcomes $=3$
$P($ Blue ball $)=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{1}{3}$
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Question 142 Marks
A bag contains a red ball , a blue ball and a yellow ball , all the balls being of the same size . Anjali takes out a ball from the bag without looking into it . what is the probability that she takes out .
(ii) red ball?
Answer
Total Number of outcomes $=3$
$
P(\text { Red ball })=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{1}{3}
$
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Question 152 Marks
A bag contains a red ball , a blue ball and a yellow ball , all the balls being of the same size . Anjali takes out a ball from the bag without looking into it . what is the probability that she takes out .
(i) Yellow ball?
Answer
Total Number of outcomes $=3$
$
P(\text { Yellow ball })=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{1}{3}
$
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Question 162 Marks
A coin is tossed twice . what is the probability of getting
(ii) atmost 1 tail
Answer
When a coin is tossed twice, the outcomes are $\{\mathrm{HH}, \mathrm{HT}, \mathrm{TH}, \mathrm{TT}\}$
Number of outcomes $=4$
Number of favourable outcomes $=3$
$
P(\text { getting atmost one tail })=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{3}{4}
$
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Question 172 Marks
A coin is tossed twice . what is the probability of getting
(i) no head
Answer
When a coin is tossed twice, the outcomes are $\{\mathrm{HH}, \mathrm{HT}, \mathrm{TH}, \mathrm{TT}\}$
Number of outcomes $=4$
Number of favourable outcomes $=1$
$
P(\text { getting no head })=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{1}{4}
$
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Question 182 Marks
Two coins are tossed together . what is the probabilty of getting different faces on the coins ?
Answer
The sample space for the experiment of tossing two coins is $\{\mathrm{HH}, \mathrm{HT}, \mathrm{TH}, \mathrm{TT}\}$
Number of outcomes $=4$
Number of favourable outcomes $=2$
$
P(\text { getting different faces on the coins })=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{2}{4}=\frac{1}{2}
$
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[2 Mark Question Answer] - Mathematics STD 10 Questions - Vidyadip