Apps. and Interpretation for IBDP Maths Ebook 1 | Page 277

78
Paper 1 – Conditional Probabilities in a Table
Example
A discrete random variable X has the following probability distribution.
x 1 2 3 4
P( X x)
�
2
5k
1
2 k 3
2 k 2
3k
(a) Find the value of k .
(b) Find P( X �4 | X � 2) .
[3]
[3]
Solution
(a) P( X �1) � P( X � 2) � P( X � 3) � P( X � 4) � 1
(b)
1 3
� � � � (A1) for substitution
2 2
2 2
5k k k 3k
1
2
8k
� 2k�1�
0
(2k�1)(4 k�1) � 0
(A1) for factorization
1
k �� (Rejected) or
2
P( X � 4 � X � 2)
P( X � 4 | X � 2) �
P( X � 2)
P( X � 4)
P( X � 4 | X � 2) �
P( X � 3) � P( X � 4)
�1
�
3�
� 4
P( X � 4 | X � 2) �
� �
3 � 1 � � 1 �
� � � 3� �
2 � 4 � � 4 �
1
k � A1 N3
4
2
2
(M1) for valid approach
(A1) for substitution
1
P( X � 4 | X � 2) � A1 N3
3
[3]
[3]
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