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A bag contains 30 balls out of which 'm' number of balls are blue in colour. a) Find the probability that a ball drawn at random from the bag is not blue. b) If 6 more blue balls are added in the bag,

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2 parts
a

30−m30\frac{30 - m}{30}

b
12

Step-by-step solution

8 steps

Here's the solution for question 29:

29. A bag contains 30 balls out of which 'm' number of balls are blue in colour.

Part a· 3 steps

Find the probability that a ball drawn at random from the bag is not blue.

  1. Identify the total number of balls and the number of blue balls.

    Total number of balls =30= 30
    Number of blue balls =m= m

  2. Calculate the number of balls that are not blue.

    Number of non-blue balls =Total balls−Number of blue balls=30−m= \text{Total balls} - \text{Number of blue balls} = 30 - m

  3. Calculate the probability of drawing a ball that is not blue.

    P(not blue)=Number of non-blue ballsTotal number of ballsP(\text{not blue}) = \frac{\text{Number of non-blue balls}}{\text{Total number of balls}} P(not blue)=30−m30P(\text{not blue}) = \frac{30 - m}{30}

    The probability that a ball drawn is not blue is 30−m30\boxed{\frac{30 - m}{30}}.

Part b12

If 6 more blue balls are added in the bag, then the probability of drawing a blue ball will be 54\frac{5}{4} times the probability of drawing a blue ball in the first case. Find the value of m.

  1. Calculate the probability of drawing a blue ball in the first case (P1(blue)P_1(\text{blue})).

    P1(blue)=Number of blue ballsTotal number of balls=m30P_1(\text{blue}) = \frac{\text{Number of blue balls}}{\text{Total number of balls}} = \frac{m}{30}

  2. Determine the new number of blue balls and total balls after adding 6 blue balls.

    New number of blue balls =m+6= m + 6
    New total number of balls =30+6=36= 30 + 6 = 36

  3. Calculate the probability of drawing a blue ball in the new case (P2(blue)P_2(\text{blue})).

    P2(blue)=New number of blue ballsNew total number of balls=m+636P_2(\text{blue}) = \frac{\text{New number of blue balls}}{\text{New total number of balls}} = \frac{m+6}{36}

  4. Set up the equation based on the given condition.

    The problem states that P2(blue)=54×P1(blue)P_2(\text{blue}) = \frac{5}{4} \times P_1(\text{blue}).

    m+636=54×m30\frac{m+6}{36} = \frac{5}{4} \times \frac{m}{30}

  5. Simplify and solve for mm.

    m+636=5m120\frac{m+6}{36} = \frac{5m}{120}

    Simplify the right side by dividing 5 and 120 by 5:

    m+636=m24\frac{m+6}{36} = \frac{m}{24}

    Cross-multiply:

    24(m+6)=36m24(m+6) = 36m 24m+144=36m24m + 144 = 36m

    Subtract 24m24m from both sides:

    144=36m−24m144 = 36m - 24m 144=12m144 = 12m

    Divide by 12:

    m=14412m = \frac{144}{12} m=12m = 12

    The value of mm is 12\boxed{12}.

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