| 1 |
P (A/B) can be evaluated by formula |
- A. <span style="color: rgb(0, 0, 0); font-family: 'Lucida Sans Unicode', 'Lucida Grande', sans-serif; font-size: 18px; line-height: 23.390625px;">P(A∩B)/P(B)</span>
- B. <span style="color: rgb(0, 0, 0); font-family: 'Lucida Sans Unicode', 'Lucida Grande', sans-serif; font-size: 18px; line-height: 23.390625px;">P(A∪B). P(B)</span>
- C. <span style="color: rgb(0, 0, 0); font-family: 'Lucida Sans Unicode', 'Lucida Grande', sans-serif; font-size: 18px; line-height: 23.390625px;">(A∪B)/P(B)</span>
- D. <span style="color: rgb(0, 0, 0); font-family: 'Lucida Sans Unicode', 'Lucida Grande', sans-serif; font-size: 18px; line-height: 23.390625px;">P(A∩B)/P(A)</span>
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| 2 |
There sets on a sofa can be occupied by four persons in. |
- A. 12 ways
- B. 7 ways
- C. 24 ways
- D. None of these
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| 3 |
"P<sub>r</sub> can be solved by the formula. |
- A. N!/ r!(n-r)!
- B. (n-r)!/r!
- C. n!(n-r!)
- D. n!(n-r)!/r!
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| 4 |
If a player well shuffles the pack of 52 playing card, then the probability of a black card form 52 playing cards is: |
- A. 1/52
- B. 13/52
- C. 26/52
- D. 4/52
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| 5 |
A coin is tossed 3 times then, then number of sample points in the sample space is: |
- A. 2<sup>3</sup>
- B. 3
- C. 8
- D. Both A & C
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| 6 |
The conditional probability P(A/B) is given by. |
- A. (A∩B)/(B)
- B. P(A∩B)/P(A)
- C. P(A∩B)/P(B)
- D. (A∩B)/P(B)
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| 7 |
If the occurance of one event is not effected by the occurance of other than these events are called |
- A. Dependent
- B. Independent
- C. Simple
- D. Compound events
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| 8 |
P(A or B) = P (A∪B) = P(A) +P(B) then A and B are. |
- A. Mutually exclusive
- B. Independent events
- C. Not mutually exclusive
- D. Dependent
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| 9 |
<sup>n</sup>P<sub>r</sub> can be solved by the formula |
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| 10 |
The probability of vowel letters form the words STATISTIC is. |
- A. 2/10
- B. 3/10
- C. 0
- D. 4/10
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