Respuesta :
The probability of outcome to be a 5 is 1/13. The probability of outcome not be 5 is 12/13. The probability of outcome to be a diamond or a king is 4/13. The probability of the outcomes to be a jack or a queen is 2/13. The probability of The outcome to be a heart and a club is 0 and the probability of the outcomes to be a card greater than 6 and less than 9 is 2/13.
A deck contains a total of 52 cards.
Total outcomes if the selected number of cards is one=52
Case A:
Each suit contain one 5 numbered card. There are four suits. Therefore, The required outcome would be 4.
Probability of an outcome is the ratio of desired outcomes to the total outcomes.
The probability of outcome to be 5, P(A):
P(A)=4/52
P(A)=1/13
Case B:
The total probability of outcomes=52/52=1
The probability of the outcome to be 5=1/13
Therefore, the probability of the outcome not to be 5, P(B):
P(B)=1-(1/13)
P(B)=12/13
Case C:
Each suit carries a king. Therefore, the probability of the outcome to be a king, P(M):
P(M)=4/52
P(M)=1/13
A suit carries 13 cards. Therefore, there will be 13 cards of diamonds.
The probability of the outcome to be a diamond, P(N):
P(N)=13/52
P(N)=1/4
The probability that the outcome is a king of diamonds, P(M[tex]\cap[/tex]N):
P(M[tex]\cap[/tex]N)=1/52
Hence, the probability that the outcome is a diamond or a king, P(M[tex]\cup[/tex]N):
P(M[tex]\cup[/tex]N)=P(M)+P(N)-P(M[tex]\cap[/tex]N)
P(M[tex]\cup[/tex]N)=(1/13)+(1/4)-(1/52)
P(M[tex]\cup[/tex]N)=16/52
P(M[tex]\cup[/tex]N)=4/13
Case D:
Each suit carries a Jack. Therefore, the probability of the outcome to be a jack, P(P):
P(P)=4/52
P(P)=1/13
Each suit carries a Queen. Therefore, the probability of the outcome to be a Queen, P(Q):
P(Q)=4/52
P(Q)=1/13
The probability that the outcome is a jack and a queen, P(P[tex]\cap[/tex]Q):
P(P[tex]\cap[/tex]Q)=0/52
P(P[tex]\cap[/tex]Q)=0
Hence, the probability that the outcome is a jack or a queen, P(P[tex]\cup[/tex]Q):
P(P[tex]\cup[/tex]Q)=P(P)+P(Q)-P(P[tex]\cap[/tex]Q)
P(P[tex]\cup[/tex]Q)=(1/13)+(1/13)-0
P(P[tex]\cup[/tex]Q)=2/13
Case E:
A card is never a heart and a club. Therefore, the probability of a card to be a heart and a club is 0.
Case F:
The numbers greater than 6 and less than 9 are the numbers between 6 and 9. The numbers between 6 and 9 are 7 and 8.
Each suit carries one 7 numbered card and one 8 numbered card. There is a total of 4 suits.
Therefore the desirable outcomes=8
The probability of the outcome required=8/52
=2/13
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