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Probability Test - 41

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Probability Test - 41
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  • Question 1
    1 / -0
    A jar of marbles contains $$4$$ blue, $$5$$ red, $$1$$ green and $$2$$ black marbles. A marble is chosen at random from the jar. After replacing it, a second marble is chosen. Find the probability of choosing a green and then a red marble. 
    Solution
    The probability of choosing green marble as first marble is $$\dfrac{1}{4+5+1+2}=\dfrac{1}{12}$$
    The green marble is replaced.
    The probability of choosing red marble as second marble is $$\dfrac{5}{4+5+1+2}=\dfrac{5}{12}$$
    The probability of choosing a green marble and then a red marble is $$\dfrac{1}{12} \times \dfrac{5}{12} = \dfrac{5}{144}$$
  • Question 2
    1 / -0
    If $$P(A)= \dfrac{1}{4}$$, $$P(B)= \dfrac{1}{2}$$, $$P(A\cup B)= \dfrac{5}{8}$$ then $$P(A\cap B)$$ is
    Solution
    Given $$P(A)=\dfrac{1}{4}, P(B)=\dfrac{1}{2}, P(A \cup B)=\dfrac{5}{8}$$.

    We have to find $$P(A \cap B)$$.

    We know that, $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

    Substitute the given values, we get

    $$\dfrac{5}{8}=\dfrac{1}{4}+\dfrac{1}{2}-P(A\cap B)$$

    $$\Rightarrow P(A\cap B)=\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{5}{8}$$

                           $$=\dfrac{2+4-5}{8}=\dfrac{1}{8}$$

    $$\therefore P(A\cap B)=\dfrac{1}{8}$$
  • Question 3
    1 / -0
    If one has three dice, what is the probability of getting three $$4's$$?

    Solution
    Three dice are rolled.
    $$n(S)=6^3=216$$.
    Let $$A$$ be the event of getting three $$4's$$.
    $$n(A)=1$$.
    Therefore, $$P(A)=\dfrac{n(A)}{n(S)}=\dfrac{1}{216}$$
    Hence, the probability of getting three $$4's$$ is $$\dfrac{1}{216}$$.
  • Question 4
    1 / -0
    Three cards are chosen at random with replacement from a standard deck of $$52$$ cards. What is the probability of choosing an ace, a spade and a four? 
    Solution
    In the deck of $$52$$ cards , there will be $$4$$ aces , $$13$$ spades and $$4$$ fours.
    The probability of selecting an ace from a deck is $$\cfrac{4}{52}=\cfrac{1}{13}$$.
    The probability of selecting a spade from a deck is $$\cfrac{13}{52}=\cfrac{1}{4}$$.
    The probability of selecting a four from a deck is $$\cfrac{4}{52}=\cfrac{1}{13}$$.
    The combined probability is $$\cfrac{1}{13} \times \cfrac{1}{4} \times \cfrac{1}{13}=\cfrac{1}{676}$$.
    So the correct option is $$B$$.
  • Question 5
    1 / -0
    A die is thrown. Let 'A' be the event that the number obtained is greater than $$3$$. Let 'B' be the event that the number obtained is less than $$5$$. Then, $$P(A\cup B)$$ is
    Solution
    A die is thrown.

    Therefore $$S=\{1,2,3,4,5,6\}$$

    $$\Rightarrow n(S)=6$$.

    Let $$A$$ be the event that the number obtained is greater than $$3$$.

    Therefore $$A=\{4,5,6\}$$

    $$\Rightarrow n(A)=3$$.

    $$\Rightarrow P(A)=\dfrac{n(A)}{n(S)}=\dfrac{3}{6}=\dfrac{1}{2}$$

    Let $$B$$ be the event that the number obtained is less than $$5$$.

    Therefore $$B=\{1,2,3,4\}$$

    $$\Rightarrow n(B)=4$$.

    $$\Rightarrow P(B)=\dfrac{n(B)}{n(S)}=\dfrac{4}{6}=\dfrac{2}{3}$$

    $$A\cap B=\{4\}$$

    $$\Rightarrow n(A\cap B)=1$$

    $$\Rightarrow P(A\cap B)=\dfrac{n(A\cap B)}{n(S)}=\dfrac{1}{6}$$

    We have to find $$P(A\cup B)$$.

    We know that $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

                                               $$=\dfrac{1}{2}+\dfrac{2}{3}-\dfrac{1}{6}$$

                                               $$=\dfrac{3+4-1}{6}$$

                                               $$=\dfrac{6}{6}$$

                                               $$=1$$

    $$\therefore P(A\cup B)=1$$
  • Question 6
    1 / -0
    An integer is chosen at random from the first $$200$$ positive integers. The probability that the integer chosen is divisible by $$6$$ or $$8$$ is 
    Solution
    The number of numbers which are divisible by $$6$$ which are less than $$200$$ is $$33$$.
    The number of numbers which are divisible by $$8$$ which are less than $$200$$ is $$25$$.
    The number of numbers which are divisible by both $$6$$ and $$8$$ which are less than $$200$$ is $$8$$.
    Therefore the number of numbers which are divisible by $$6$$ or $$8$$ is $$33+25-8=50$$.
    The probability is $$\cfrac{50}{200}=\cfrac{1}{4}$$.
    The correct option is $$A$$.
  • Question 7
    1 / -0
    If $$P(A)= \dfrac{3}{5}$$ and $$P(B)= \dfrac{2}{3}$$ and A and B are indepedent events, then which of the following is incorrect?
    Solution
    Given $$P(A)=\frac{3}{5}$$ and $$P(B)=\frac{2}{3}$$
    $$P(A\cap \overline { B })=P(A)P(\overline { B } )=\frac{1}{5} < \frac{1}{3} $$
    $$P(A\cup B) = P(A)+P(B)-P(A)P(B)=\frac{1}{5}+\frac{2}{3}>\frac{2}{3}$$
    $$P(A\cap B)=P(A)P(B)=\frac{2}{5}$$ 
    $$\Rightarrow \frac{4}{15} \le P(A\cap B) \le \frac{3}{5}$$
    Therefore option $$A$$ is correct
  • Question 8
    1 / -0
    In the experiment of rolling a dice, find the probability of getting an even number that is multiple of 3.
    Solution
    The number of ways we can get even number in a throw is $$3$$
    Among those $$3$$ , the number of numbers which are divisible by $$3$$ is only $$1$$
    therefore probability is $$\dfrac{1}{6}$$
    So the correct option is $$D$$
  • Question 9
    1 / -0
    While rolling a fair die, find the probability of event that the number that turns in place is even as well as divisible by four.
    Solution
    A fair die is thrown.
    We know $$ S=\{1,2,3,4,5,6\}$$
    Thus $$ n(S)=6$$
    Let $$A$$ be the event of getting a number that is even as well as divisible by $$4$$.
    $$\Rightarrow A=\{4\}$$
    $$\Rightarrow  n(A)=1$$
    Hence, the probability of event that the number that turns in place is even as well as divisible by four is $$P(A)=\dfrac{n(A)}{n(S)}$$.
    $$\Rightarrow P(A)=\dfrac{1}{6}$$
    Thus the probability of event that the number that turns in place is even as well as divisible by four is $$\dfrac{1}{6}$$.
  • Question 10
    1 / -0
    What is the probability that the wheel stops at red or pink?

    Solution
    Totally there are $$8$$ parts in the wheel.
    Therefore $$n(S)=8$$.
    Let $$A$$ be the event that the wheel stops at red.
    $$n(A)=2$$.
    $$\Rightarrow P(A)=\dfrac{n(A)}{n(S)}=\dfrac{2}{8}=\dfrac{1}{4}$$
    Let $$B$$ be the event that the wheel stops at pink.
    $$n(B)=1$$.
    $$\Rightarrow P(B)=\dfrac{n(B)}{n(S)}=\dfrac{1}{8}$$
    We have to find the probability that the wheel stops at red or pink. 
    That is we have to find $$P(A\cup B)$$.
    Since $$A$$ and $$B$$ are mutually exclusive events, we can say that $$P(A\cup B)=P(A)+P(B)$$
    $$\Rightarrow P(A\cup B)=\dfrac{1}{4}+\dfrac{1}{8}$$
    $$=\dfrac{2+1}{8}$$
    $$\therefore P(A\cup B)=\dfrac{3}{8}$$
    Hence, the probability that the wheel stops at red or pink is $$\dfrac{3}{8}$$.
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