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X, is assumed to be a random variable by placing letters of the word “YACHT” in a hat. If these letters are withdrawn then it gets replaced. A value of 1 is given when a vowel is retrieved, and a value of 2 is given when a consonant is retrieved while a value of 3 is given to all other retrieved letters. Construct a probability distribution of X Is this correct: Letter, X Y A C H T Probability, PX 4/5 1/5 4/5 4/5 4/5

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1 is obtained when X attains the value of A, which occurs 1 in 5 times. Similarly, 2 is obtained when X attains any of Y,T,C or H, which occurs 4 in 5 times. Note that 3 is never obtained as it X can be either vowel or a consonant(only these two possibilities) ,So 3 never occurs(or we can say it occurs...
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1 is obtained when X attains the value of A, which occurs 1 in 5 times. Similarly, 2 is obtained when X attains any of Y,T,C or H, which occurs 4 in 5 times. Note that 3 is never obtained as it X can be either vowel or a consonant(only these two possibilities) ,So 3 never occurs(or we can say it occurs 0 times in 5). Therefore, the probability distribution for X is as follows: P(1) = P( X = A) = 1/5 P(2) = P(X = Y, C, H, T) = 4/5 P(3) = P(X != YACHT) = 0 read less
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There are 3 values- {1, 2, 3} Since vowels (value 1), consonants (value 2), and others i.e. y (value 3). So I can only assume that Y is considered "other", and not a vowel or consonant. So we get: Letter - X Y - 3 A - 1 C - 2 H - 2 T - 2 Hence, P(X=1) = 1/5 P(X=2) = 3/5 P(X=3) =...
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Since Y-2, A-1, C-2, H-2, T-2 so we can write Sample space={ 1 and 2}. This is write and the above one you mentioned is also write. Because this contains only a vowel and others are consonants and this does not contain any sample space that is 3.
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Well Distribution should be be in terms of random numbers which are S = {1,2,3} P(1) = P( X = A) = 1/5 P(2) = P(X = Y, C, H, T) = 4/5 P(3) = P(X != YACHT) = 0
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M.Sc.B.Ed. Csir-net, set, ctet( mathematics)

Random variable X can take the values 1,2. P(getting vowel) = P(X=1) = 1/5 P(getting consonant) = P(X=2)= 4/5 X 1 2 P(X) 1/5 2/5
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Since vowel values 1 ,consonant values 2 and others value 3 so I assume that 'y' as other and therefore we can write this like XY-3A-1C-2H-2T-2..henceforth P(1)=P=A=1/5;P(2)=P (X=Y,C,H,T)=4/5;P (3)=(X!=YACHT)=0
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Mathematics Teacher

NOT CURRECT. THERE ARE TWO PROPERTIES OF PROBABILITY DISTRIBUTION FUNCTION. f(x) greater or equal to zero . ,sum of f(xi) equal to 1.
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Since Y-2, A-1, C-2, H-2, T-2 Sample space = {1 and 2}. Yes
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M.Sc.B.Ed. Csir-net, set, ctet( mathematics)

Here....
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