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Probability Test - 37

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Probability Test - 37
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  • Question 1
    1 / -0
    Consider the word $$W = MISSISSIPPI$$ Number of ways in which the letters of the word W can be arranged if at least one vowel is separated from rest of the vowels
    Solution
    W = MISSISSIPPI , Letters= 11
    1 M. 4I's, 4S's, 2P's
    Total number of arranging these letters
    $$=\quad \cfrac { 11! }{ 4!4!2! } $$
    Suppose all I's are together , then the Number of ways 
    $$=\quad \cfrac { \left( 1+1+4+2 \right) ! }{ 4!2! } \quad =\quad \cfrac { 8! }{ 4!2! } $$
    $$\therefore$$ Number of ways in which atleast one vowel is separated
    $$=\quad \cfrac { 11! }{ 4!4!2! } \quad -\quad \cfrac { 8! }{ 4!2! } \\ =\quad \cfrac { 8!16! }{ 4!4!2! } $$
  • Question 2
    1 / -0
    From a well shuffled pack of $$52$$ playing cards, four are drawn at random. The probability that all are spades, but one is a king is: 
    Solution

    Total No. Of cards $$=52$$

    Total cards of spade $$=13$$

    $$Probability =\dfrac{ Conditional\ case} {total\ case}$$


    So, 

    Total no. Of conditional case of drawing $$4$$ cards of spade but one is king  $$=$$ no. of case of selecting all four cards of spade suit 

    $$=\ ^{12}C_3\times 1$$

    Total no. of case $$=\ ^{52}C_4$$

    Probability of being one card as king out of four $$=\dfrac{^{12}C_3}{ ^{52}C_4}$$

  • Question 3
    1 / -0
    A box contains $$10$$ items, $$3$$ of which are defective. If $$4$$ are selected at random without replacement, the probability that at least $$2$$ are defective is?
    Solution

  • Question 4
    1 / -0
    There are six letters $$L_1, L_2, L_3, L_4, L_5, L_6$$ are their corresponding six envelopes $$E_1, E_2, E_3, E_4, E_5, E_6$$. Letters having odd value can be put into odd value envelopes and even value letters can be put into even value envelopes, so that no letter go into the right envelopes, then number of arrangement equals?
    Solution
    There are three odd envelops and three even envelops 
    Favourable ways so that  letters having odd value can be put into odd value envelopes and even value letters can be put into even value envelopes, so that no letter go into the right envelopes are
    For odd 1,3,5 -_,_,_
    5 1 3,3 5 1
    For even 2,4,6- _,_,_
    6 2 4,4 6 2
    There are four ways
  • Question 5
    1 / -0
    If $$C$$ and $$D$$ are two events such that $$C\subset  D$$ and $$P(D)\ne 0$$, then the correct statement among the following is
    Solution
    $$P(C|D)=\cfrac{P(C\cap D)}{P(D)}=\cfrac{P(C)}{P(D)}$$
    and $$0< P(D)\le 1$$
    $$\cfrac{P(C)}{P(D)}\ge P(C)$$
  • Question 6
    1 / -0
    Read the passage given carefully before attempting these questions.
    A standard deck of playing cards has $$52$$ cards. There are four suit (clubs, diamonds, hearts and spades), each of which has thirteen numbered cards ($$2, ., 9, 10$$, Jack, Queen, King, Ace)
    In a game of card, each card is worth an amount of points. Each numbered card is worth its number (e.g., a $$5$$ is worth $$5$$ points); the Jack, Queen and King are each worth $$10$$ points; and the Ace is worth your choice of "either $$1$$ point or $$11$$ points". The object of the game is to have more points in your set of cards than your opponent without going over $$21$$. Any set of cards with sum greater than $$21$$ automatically loses.
    Here's how the game is played. You and your opponent are each dealt two cards. Usually the first card for each player is dealt face down, and the second card for each player is dealt face up. After the initial cards are dealt, the first player has the option of asking for another card or not taking any cards. The first player can keep asking for more cards until either he or she goes over $$21$$, in which case the player loses, or stops at some number less than or equal to $$21$$. When the first player stops at some number less than or equal to $$21$$, the second player then can take more cards until matching or exceeding the first player's number without going over $$21$$, in which case the second player wins, or until going over $$21$$, in which case the first player wins.
    We are going to simplify the game a little and assume that all cards are dealt face up, so that all cards are visible. Assume your opponent is dealt cards and plays first.
    The chance that the second card will be a heart and a Jack, is?
  • Question 7
    1 / -0
    Let A and B be two events such that $$P(\overline{A\cup B})=\dfrac{1}{6}, P(A\cap B)=\dfrac{1}{4}$$ and $$P(\overline{A})=\dfrac{1}{4}$$, where $$\overline{A}$$ stands for the complement of the event A. Then the events A and B are?
    Solution
    P(A')=1/4 so P(A)=3/4
    P(AUB)'=1/6, P(AUB)=5/6
    P(AB)=1/4
    P(AUB)=P(A)+P(B)-P(AB)=3/4+P(B)-1/4
    5/6=1/2+P(B)
    P(B)=2/6=1/3
    P(A)*P(B)=1/3*3/4=1/4=P(AB)
    A and B are independent but not equally likely 
  • Question 8
    1 / -0
    A card is down & replaced in ordinary pack of 52 playing cards. Minimum number of times must a card be drawn so that there is atleast an even chance of drawing a heart, is
    Solution
    Probability of drawing a heat in one draw $$=\dfrac{1}{4}$$
    For $$n$$ drawing Probability of not getting any heat $$=\dfrac{3}{4}$$
    $$\therefore $$ Probability of drawing at least one heat
    $$1-(\dfrac{3}{4})^n$$
    By addition
    $$1-(\dfrac{3}{4})^n \ge \dfrac{1}{2}$$
    $$\Rightarrow (\dfrac{3}{4})^n \le \dfrac{1}{2}$$
    $$\Rightarrow (log \dfrac{3}{4})\le log\dfrac{1}{2}$$
    $$\Rightarrow n \le \dfrac{log\dfrac{1}{2}}{log\dfrac{3}{4}}$$
    $$\Rightarrow n\le 2.4$$
    $$\therefore \boxed{n=2}$$

  • Question 9
    1 / -0
    A box contains $$7$$ tickets, numbered from $$1$$ to $$7$$ inclusive. If $$3$$ tickets are drawn from the box without replacement, one at a time, determine the probability that they alternatively either odd-even-odd or even -odd - even.
    Solution
    For odd-even-odd=
    $$P(A)=\dfrac{4}{7}\times \dfrac{3}{6}\times \dfrac{3}{5}=\dfrac{6}{35}$$
    For even-odd-even=
    $$P(B)=\dfrac{3}{7}\times \dfrac{4}{6}\times \dfrac{2}{5}=\dfrac{4}{35}$$
    $$\therefore P=P(A)+P(B)=\dfrac{6}{35}+\dfrac{4}{35}=\dfrac{10}{35}=\dfrac{2}{7}$$
  • Question 10
    1 / -0
    Given two independent events A and B such that $$P\left( A \right) = 0.3,$$ $$P\left( A\cap B \right) = 0.1$$. Find $$P\left( {A \cap B'} \right)$$
    Solution
    Given $$P\left(A\right)=0.3,\,\,P\left(A\cap\,B\right)=0.1$$

    If $$A$$ and $$B$$ are independent events, $$P\left(A\cap\,B\right)=P\left(A\right)P\left(B\right)$$
    $$\Rightarrow\,P\left(B\right)=\dfrac{P\left(A\cap\,B\right)}{P\left(A\right)}=\dfrac{0.1}{0.3}=\dfrac{1}{3}$$

    If $$A$$ and $$B$$ are independent events,
    $$P\left(A\cap{B}^{\prime}\right)=P\left(A\right)\times P\left({B}^{\prime}\right)$$
    $$=P\left(A\right)\left(1-P\left(B\right)\right)$$
    $$=0.3\left(1-\dfrac{1}{3}\right)$$
    $$=0.3\times\dfrac{3-1}{3}=0.3\times\dfrac{2}{3}=0.1\times 2=0.2$$
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