In a simultaneous throw of 2 dice,find P(A te | vedclass

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(A) The total number of outcomes when throwing $2$ dice is $6 	imes 6 = 36$.Event $A$: Getting a total of $11$.The favorable outcomes are $(5, 6)$ an

Vedclass · Accepted Answer

$\frac{11}{36}$

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(A) The total number of outcomes when throwing $2$ dice is $6 	imes 6 = 36$.Event $A$: Getting a total of $11$.The favorable outcomes are $(5, 6)$ and $(6, 5)$.Thus,$n(A) = 2$,and $P(A) = \frac{2}{36}$.Event $B$: Getting an odd number on each die.The odd numbers on a die are ${1, 3, 5}$.The favorable outcomes are $(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)$.Thus,$n(B) = 9$,and $P(B) = \frac{9}{36}$.Intersection $A \cap B$: Getting a total of $11$ $AND$ an odd number on each die.Since the sum of two odd numbers is always even,it is impossible to get a total of $11$ (which is odd) if both dice show odd numbers.Thus,$A \cap B = \emptyset$,and $P(A \cap B) = 0$.Using the addition rule of probability:$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$P(A \cup B) = \frac{2}{36} + \frac{9}{36} - 0 = \frac{11}{36}$.

In a simultaneous throw of $2$ dice,find $P(A \text{ or } B)$ if $A$ denotes the event 'a total of $11$' and $B$ denotes the event 'an odd number on each die'.

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