Explanation
Solution
1. Basic Information:
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Total faces on the die =6
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Outcomes on the die: {−2,−1,0,1,2,3}
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Total outcomes in 5 throws: 65=7776
2. Product Positive hone ki Condition:
Product tabhi positive hoga jab:
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Kisi bhi throw mein 0 na aaye (kyunki 0 aane se product 0 ho jayega).
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Negative numbers (−2,−1) ki sankhya even (0,2,4) honi chahiye.
Let:
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P = Positive numbers aane ki probability {1,2,3}=63=21
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N = Negative numbers aane ki probability {−2,−1}=62=31
3. Cases Calculate karna:
Hame 5 throws mein even number of negatives chahiye:
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Case 1: 0 Negatives (Saare Positive)
5C0×(N)0×(P)5=1×(31)0×(21)5=321
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Case 2: 2 Negatives (aur 3 Positive)
5C2×(N)2×(P)3=10×(31)2×(21)3=10×91×81=7210=365
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Case 3: 4 Negatives (aur 1 Positive)
5C4×(N)4×(P)1=5×(31)4×(21)1=5×811×21=1625
4. Total Probability:
P(\text{Product} > 0) = \frac{1}{32} + \frac{5}{36} + \frac{5}{162}
L.C.M lene par (32,36,162 ka L.C.M 2592 hai):
Correct Option: (4)
