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Statistics

Notes Compare P’ – P’ with the properties of (unconditiond) probabilities given in Sec. 6.2.2. You will
1 5
find that the conditional probabilities, given the event H, have all the properties of unconditional
probabilities, which are sometimes called the absolute properties.
We can use the conditional probatii:itics to compute the unconditional probabilities of events
by employing the following obvious fact,

P(A  H) = P(H) P(A  H) ... (10)
obtained from Definition 3 of P(A  H).
Here is an important remark related to (10).

Notes See E 14 for the interpretations of P(A | H) and P(A | H)
C

Remark 3 : Relation (10) holds even if P(H) = 0, provided we interpret P(A | H) = 0 if P(H) = 0. In
words, this means that if the probability of -occurrence of H is zero, we say that the probability
of occurrence of A, given that H has occurred, is also zero. This is so, because P(H) = 0 implies
P(A  H) = 0, (A H) being a subset of H,

We now give an example to illustrate the use of Relation (10).

Example 14: Two cards are drawn at random and without ieplacement from a pack of 52
playing cards. Let us find the probability that both the cards are red.
Let A and A denote, respectively the events that cards drawn on the first and second draw are
1 2
red. Then by the classical definition, P(A ) = 26/52, since there are 26 red cards. If the first card is
1
red, we are left with 25 red cards in the pack of 51 cards. Hence P(A | A ) = 25/51. Thus, the
2 1
probability P(A  A ) of both cards being red is
1 2
P(A  A ) = P(A )P(A | A )
1 2 1 2 1
26 25
= ,  0.245.
52 51
Relation (10) specifies the probability of A  H in terms of P(H) and P(A/H). We can extend this
relation to obtain the probability, P(A  A  A ) in terms of P(A ), P(A | A ) and P(A | A 
1 2 3 1 2 1 3 1
A ). We, of course, assume that P(A ) and P(A  A ) are both positive. Can you guess what this
2 1 1 2
relation could be? Suppose we write
P(A  A ) P(A  A  A )
P(A  A  A ) = P(A ) 1 2 . 1 2 3
1 2 3 1 P(A ) P(A  A )
1 1 2
Does this give you any clue? This gives us,

P(A A A ) = P(A ) . P(A | A ) . P(A | A A ).
1 2 3 1 2 1 3 1 2
Now let us use this to compute some probabilities.

Example 15: A box of mangoes is inspected by examining three randomly selected
mangoes drawn without replacement. If all the three mangoes are good, the box is sent to the
market, otherwise it is rejected. Let us calculate the probability that a box of 100 mangoes
containing 90 good mangoes and 10 bad ones will pass the inspection.

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