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A coin tossed three times, find the probability of getting: (i) 3 heads (ii) Not 3 heads (iii) At least 2 heads (iv) At least two tails

Mathematics

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4 parts
i

18\frac{1}{8}

ii

78\frac{7}{8}

iii

12\frac{1}{2}

iv

12\frac{1}{2}

15\frac{1}{5}

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Here are the solutions for the Mathematics questions.

SUBJECT: MATHEMATICS

1. A coin tossed three times, find the probability of getting:

Step 1: Determine the sample space.
When a coin is tossed three times, the total number of possible outcomes is 23=82^3 = 8.
The sample space SS is:

S={HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}

Part i

3 heads

The event of getting 3 heads is E1={HHH}E_1 = \{HHH\}.
The number of outcomes in E1E_1 is 1.

P(3 heads)=Number of outcomes in E1Total number of outcomes=18P(3 \text{ heads}) = \frac{\text{Number of outcomes in } E_1}{\text{Total number of outcomes}} = \frac{1}{8}

The probability of getting 3 heads is 18\boxed{\frac{1}{8}}.

Part ii

78\frac{7}{8}

Not 3 heads

This is the complement of getting 3 heads.

P(Not 3 heads)=1−P(3 heads)=1−18=78P(\text{Not 3 heads}) = 1 - P(3 \text{ heads}) = 1 - \frac{1}{8} = \frac{7}{8}

The probability of not getting 3 heads is 78\boxed{\frac{7}{8}}.

Part iii

12\frac{1}{2}

At least 2 heads

The event of getting at least 2 heads means getting 2 heads or 3 heads.
Outcomes with 2 heads: {HHT,HTH,THH}\{HHT, HTH, THH\}
Outcomes with 3 heads: {HHH}\{HHH\}
The event E3={HHT,HTH,THH,HHH}E_3 = \{HHT, HTH, THH, HHH\}.
The number of outcomes in E3E_3 is 4.

P(At least 2 heads)=Number of outcomes in E3Total number of outcomes=48=12P(\text{At least 2 heads}) = \frac{\text{Number of outcomes in } E_3}{\text{Total number of outcomes}} = \frac{4}{8} = \frac{1}{2}

The probability of getting at least 2 heads is 12\boxed{\frac{1}{2}}.

Part iv

12\frac{1}{2}

At least two tails

The event of getting at least two tails means getting 2 tails or 3 tails.
Outcomes with 2 tails: {HTT,THT,TTH}\{HTT, THT, TTH\}
Outcomes with 3 tails: {TTT}\{TTT\}
The event E4={HTT,THT,TTH,TTT}E_4 = \{HTT, THT, TTH, TTT\}.
The number of outcomes in E4E_4 is 4.

P(At least two tails)=Number of outcomes in E4Total number of outcomes=48=12P(\text{At least two tails}) = \frac{\text{Number of outcomes in } E_4}{\text{Total number of outcomes}} = \frac{4}{8} = \frac{1}{2}

The probability of getting at least two tails is 12\boxed{\frac{1}{2}}.

2. The favorite fruits of 40 students are in the table below:

FruitOrangesApplesMangoesGrapes
Frequency157810

  1. Calculate the angles for the pie chart.

    The total number of students is 40. The total angle in a circle is 360∘360^\circ.
    Angle for Oranges:

    1540×360∘=135∘\frac{15}{40} \times 360^\circ = 135^\circ

    Angle for Apples:

    740×360∘=63∘\frac{7}{40} \times 360^\circ = 63^\circ

    Angle for Mangoes:

    840×360∘=72∘\frac{8}{40} \times 360^\circ = 72^\circ

    Angle for Grapes:

    1040×360∘=90∘\frac{10}{40} \times 360^\circ = 90^\circ

    Note: I cannot draw the pie chart in this format, but these are the calculations needed to construct it.

  2. Calculate the fraction of mangoes.

    The frequency of mangoes is 8.
    The total number of students is 40.

    Fraction of mangoes=Frequency of MangoesTotal number of students=840=15\text{Fraction of mangoes} = \frac{\text{Frequency of Mangoes}}{\text{Total number of students}} = \frac{8}{40} = \frac{1}{5}

    The fraction of mangoes is 15\boxed{\frac{1}{5}}.

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