More about probability - (15.4-15.5)

Recognise the concept and notation of conditional probability

  • Theory
  • Examples
    It is found that in a set of 2 free throws of a basketball player, the performance
    in the second shot is affected by the first shot.
    The probability for him to have the first successful free throw is 0.86, while the
    probability for him to have both successful free throws is 0.52.

    Find the probability for him to have the second successful free throw, given that he
    has the first successful free throw.
  • Solutions
    Let A be the event that the player has the first successful free throw,
    B be the event that the player has the second successful free throw.

    \(\because \begin{array}{1}P(A) = 0.86\\P(A \cap B) = {0.52^{}}\end{array}\)

    \( \begin{array}{1}\therefore \qquad P(B|A) = & \,\frac{{P(A \cap B)}}{{P(A)}}\\ = & \,\frac{{0.52}}{{0.86}}\\ = & \,{\underline{\underline {\frac{{26}}{{43}}}} ^{}}\end{array}\)

Use permutation and combination to solve problems relating to probability

  • Theory
  • Examples
    a) There are 30 students in S5A, 28 students in S5B and 24 students in S5C.
    If two students are selected randomly, find the probability of two students
    from the same class are selected.

    b)There are 9 digital cameras, A, B, C, D, E, F, G, H and I.
    If these 9 digital cameras are randomly arranged in a row, find the probability
    that D, E, F and G are placed next to each other.
  • Solutions

    a) Number of combinations of selecting two students from the same class
    \(\begin{array}{l} = C_2^{30} + C_2^{28} + C_2^{24}\\ = 435 + 378 + {276^{}}\\ = 1{\rm{ }}{089^{}}\end{array}\)
    Number of combinations of selecting two students
    \(\begin{array}{l} = C_2^{82}\\ = 3{\rm{ }}{321^{}}\end{array}\)

    P(Two students from the same class are selected)
    \(\begin{array}{l} = \frac{{1{\rm{ }}089}}{{3{\rm{ }}321}}\\ = {\underline{\underline {\frac{{121}}{{369}}}} ^{}}\end{array}\)

    b) P(D, E, F and G are placed next to each other)
    \(\begin{array}{l} = \frac{{6{\rm{ }}!\; \times 4{\rm{ }}!}}{{9{\rm{ }}!}}\\ = {\underline{\underline {\frac{1}{{21}}}} ^{}}\end{array}\)