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Maths 3 Marks Question

Probability · Gujarat Board (GSEB) STD 10 (GSEB - English Medium). 8 questions.

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Question 13 Marks
A box contains $5$ red marbles, $8$ white marbles and $4$ green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
  1. red $?$
  2. white $?$
  3. not green $?$
Answer
Total number of marbles in the box $= 5 + 8 + 4 = 17$
$\therefore$Total number of elementary events $= 17$ $Proabibilty\;of\;the\;event=\frac{Number\;of\;favourble\;outcomes}{Total\;number\;of\;possible\;outcomes}$
  1. There are $5$ red marbles in the box.
    $\therefore$Favourable number of elementary events $= 5$
    $\therefore P ($getting a red marble$) = \frac5{17}$
  2. There are 8 white marbles in the box.
    $\therefore$Favourable number of elementary events $= 8$
    $\therefore P ($getting a white marble$) = \frac8{17}$
  3. There are $5 + 8 = 13$ marbles in the box, which are not green.
    $\therefore$ Favourable number of elementary events $= 13$
    $\therefore P ($not getting a green marble$) =\frac{13}{17}$
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Question 23 Marks
  1. Two dice, one blue and one grey, are thrown at the same time. Complete the following table:
    Event: Sum on $2$ dice $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $11$ $12$
    Probability $\frac{1}{{36}}$           $\frac{5}{{36}}$       $\frac{1}{{36}}$
  2. A student argues that there are $11$ possible outcomes $2, 3, 4, 5, 6, 7, 8, 9, 10, 11$ and $12.$ Therefore, each of them has a probability $\frac{1}{11}$. Do you agree with this argument? Justify your answer.
Answer
  1. It can be observed that,
    To get the sum as $2,$ possible outcomes $= (1, 1)$
    To get the sum as $3,$ possible outcomes $= (2, 1)$ and $(1, 2)$
    To get the sum as $4,$ possible outcomes $= (3, 1), (1, 3), (2, 2)$
    To get the sum as $5,$ possible outcomes $= (4, 1), (1, 4), (2, 3), (3, 2)$
    To get the sum as $6,$ possible outcomes $= (5, 1), (1, 5), (2, 4), (4, 2), (3, 3)$
    To get the sum as $7,$ possible outcomes $= (6, 1), (1, 6), (2, 5), (5, 2), (3, 4), (4, 3)$
    To get the sum as $8,$ possible outcomes $= (6, 2), (2, 6), (3, 5), (5, 3), (4, 4)$
    To get the sum as $9,$ possible outcomes $= (3, 6), (6, 3), (4, 5), (5, 4)$
    To get the sum as $10,$ possible outcomes $= (4, 6), (6, 4), (5, 5)$
    To get the sum as $11,$ possible outcomes $= (5, 6), (6, 5)$
    To get the sum as $12,$ possible outcomes $= (6, 6)$
    Event $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $11$ $12$
    Probability $\frac{1}{{36}}$ $\frac{2}{{36}}$ $\frac{3}{{36}}$ $\frac{4}{{36}}$ $\frac{5}{{36}}$ $\frac{6}{{36}}$ $\frac{5}{{36}}$ $\frac{4}{{36}}$ $\frac{3}{{36}}$ $\frac{2}{{36}}$ $\frac{1}{{36}}$
  2. The probability of each of these sums will not be $\frac{1}{11}$ as their sums are not equally likely.
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Question 33 Marks
One card is drawn from a well-shuffled deck of $52$ cards. Find the probability of getting
  1. a king of red colour
  2. a face card
  3. a red face card
  4. the jack of hearts
  5. a spade
  6. the queen of diamonds
Answer
Total number of cards in one deck of cards is $52.$
$\therefore$ Total number of outcomes $n = 52$
  1. Let $E_1= $Event of getting a king of red color. So number of outcomes favourable to $E_1 m = 2$ So $P(E_1)=\frac mn=\frac{2}{52}=\frac{1}{26}$
  2. Let $E_2=$ Event of getting a face card
    $\therefore$ Numbers of outcomes favourable to $E_2, m= 12. $ Hence $P(E_2) =\frac mn$= $\frac{{12}}{{52}} = \frac{3}{{13}}$
  3. Let $E_3=$ Event of getting a red face card
    $\therefore$ Numbers of outcomes favourable to $E_3= 6$ [$\because$ there are 6 red face cards in a deck ] Hence $P(E_3) = \frac mn =\frac{6}{52}=\frac{3}{26}$
  4. Let $E_4=$ Event of getting a jack of heart
    $\therefore$ Numbers of outcomes favourable to $E_4= 1$ [$\because$ there is only one jack of heart in deck of cards.]
    Hence $P(E_4)=\frac mn =$ $\frac{1}{{52}}$
  5. Let $E_5=$ Event of getting a spade
    $\therefore$ Numbers of outcomes favourable to $E_5= 13$ [$\because$ there are 13 spade in a deck]
    Hence $P(E_5) =\frac mn =$ $\frac{{13}}{{52}}$
  6. Let $E_6=$ Event of getting the queen of diamond
    $\therefore$ Numbers of outcomes favourable to $E_6= 1$ [$\because$ there is only one queen of diamond in a deck]
    Hence, $P(E_6)= \frac mn =  \frac{1}{{52}}$
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Question 43 Marks
A die is thrown once. Find the probability of getting
  1. a prime number;
  2. a number lying between $2$ and $6$
  3. an odd number.
Answer
Number of all possible outcome $(1, 2, 3, 4, 5, 6) = 6$
  1. Let $E$ be the event of getting a prime number. Then, the outcomes favourable to $E$ are $2, 3$ and $5.$
    Therefore, the number outcomes favourable to $E$ is $3.$
    So, $\mathrm { P } ( \mathrm { E } ) = \frac { \text { Number of outcomes favourable to } \mathrm { E } } { \text { Number of all possible outcomes } } = \frac { 3 } { 6 } = \frac { 1 } { 2 }$
  2. Let $E$ be the event of getting a number lying between $2$ and $6.$
    Then, the outcomes favourable to $E$ are $3, 4$, and $5$. Therefore, the number of outcomes favourable to $E$ is $3.$
    So, $\mathrm { P } ( \mathrm { E } ) = \frac { \mathrm { Number~of~outcomes } \text { favourable to } \mathrm { E } } { \text { Number of all possible outcomes } } = \frac { 3 } { 6 } = \frac { 1 } { 2 }$
  3. Let $E$ be the event getting an odd number.
    Then, the outcomes favourable to $E$ are $1, 3,$ and $5.$ Therefore, the number of outcomes favourable to $E$ is $3.$
    So, $\mathrm { P } ( \mathrm { E } ) = \frac { \text { Number of outcomes favourable to } \mathrm { E } } { \text { Number of all possible outcomes } } = \frac { 3 } { 6 } = \frac { 1 } { 2 }$
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Question 53 Marks
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers $1, 2, 3, 4, 5, 6, 7, 8 ($see figure$)$ and these are equally likely outcomes. What is the probability that it will point at:
  1. $8?$
  2. an odd number$?$
  3. a number greater than $2?$
  4. a number less than $9?$
Answer
Out of $8$ numbers, an arrow can point any of the numbers in $8$ ways.
$\therefore$Total number of outcomes $= 8$
$Proabibilty\;of\;the\;event=\frac{Number\;of\;favourble\;outcomes}{Total\;number\;of\;possible\;outcomes}$
  1. Favourable number of outcomes $= 1$
    Hence, $P ($arrow points at $8) = \frac18$
  2. Favourable number of outcomes $= 4$
    Hence, $P ($arrow points at an odd number$) = \frac48=\frac12$
  3. Favourable number of outcomes $= 6$
    Hence, $P ($arrow points at a number greater than $ 2) = \frac68=\frac34$
  4. Favourable number of outcomes $= 8$
    Hence, $P ($arrow points at a number less than $ 9) = \frac88=1$
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Question 63 Marks
A piggy bank contains hundred $50$ p coins, fifty $₹\ 1$ coins, twenty $₹\ 2$ coins and ten $₹\ 5$ coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin
  1. will be a $50$ p coin$?$
  2. will not be a $₹\ 5$ coin$?$
Answer
Number $50\ p$ coins in the piggy bank $= 100$
Number of $Re. 1$ coins in the piggy bank $= 50$
Number of $Rs. 2$ coins in the piggy bank $= 20$
Number of $Rs. 5$ coins in the piggy bank $= 10$
∴ Total number of coins in the piggy bank $= 100 + 50 + 20 -10 = 180$
∴ Number of all possible outcomes $= 180$
  1. Number of favourable outcomes to the event that the coin will be a $50\ p$ coin $= 100$
    $∴$ Probability that the coin will be a $50\ p$ coin
    $\frac{Number\;of\;favourable\;outcomes\;to\;the\;event\;that\;the\;coin\;will\;be\;a\;50\;p\;coin}{\;Number\;of\;all\;possible\;outcomes}$ $\frac{100}{180}=\frac59$
  2. Number of favourable outcomes to the event that the coin will not be a $Rs. 5$ coin
    $= 100 + 50 + 20 = 170$
    $∴$ Probability that the coin will not be $Rs. 5$ coin
    $\frac{Total \ no.\;of\;favourable\;outcomes}{Total\;number\;of\;possible\;outcomes}$ $\frac{170}{180}=\frac{17}{18}$
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Question 73 Marks
Harpreet tosses two different coins simultaneously $($say, one is of $₹\ 1$ and other of $₹\ 2).$ What is the probability that he gets at least one head$?$
Answer
If we suppose , $H$ for 'head' and $T$ for 'tail'. When two coins are tossed simultaneously, the possible outcomes are $(H,H),(H,T),(T,H),(T,T),$ which are all equally likely. Here $(Rs.2).$ Similarly $(H,T)$ means head up on the first coin $($say on $Rs.1)$ and head up on the second coin and so on.
The outcomes favourable to the event $E,$ 'at least one head' are $(H, H),(H, T)$ and $(T, H).$
So, the number of outcomes favourable to $E$ is $3.$
Therefore, $P(E) = \frac{3}{4}$
i.e, the probability that Harpreet gets at least one head is $\frac{3}{4}$.
Note you can also find $P(E)$ as follows:
$P(E) = 1 - P ( \overline { E } )$ = 1 - $\frac{1}{4}$= $\frac{3}{4} ($Since $P ( \overline { E } ) = P($no head$) = \frac{1}{4}$
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Question 83 Marks
Two dice, one blue and one grey are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is
  1. $8?$
  2. $13?$
  3. less than or equal to $12?$
Answer
When the blue die shows $‘1’,$ the grey die could show any one of the numbers $1, 2, 3, 4, 5, 6.$ The same is true when the blue die shows $‘2’, ‘3’, ‘4’, ‘5’$ or $‘6’$. The possible outcomes of the experiment are listed in the table below; the first number in each ordered pair is the number appearing on the blue die and the second number is that on the grey die.

So, the number of possible outcomes $= 6 × 6 = 36.$
  1. The outcomes favourable to the event ‘the sum of the two numbers is $8’$ denoted by $E,$ are: $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2) ($see Fig.$)$
    i.e., the number of outcomes favourable to $E = 5.$
    Hence, $P(E) = \frac{5}{36}$
  2. As you can see from Fig., there is no outcome favourable to the event $F,$ ‘the sum of two numbers is $13’.$
    So, $P(F) = \frac{0}{36} = 0$
  3. As you can see from Fig., all the outcomes are favourable to the event $G$, ‘sum of two numbers $≤ 12’.$
    So, $P(G) = \frac{36}{36}=1$
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