Question
Using truth table prove that : $\sim p \wedge q=(p \vee q) \wedge \sim p$
| I | II | III | IV | V | VI |
| p | q | $\sim p$ | $\sim p \wedge q$ | $p \vee q$ | $(p \vee q) \wedge \sim p$ |
| T | T | F | F | T | F |
| T | F | F | F | T | F |
| F | T | T | T | T | T |
| F | F | T | F | F | F |
From column (IV) and (VI), we get
∴ ∼p ˄ q ≡ (p ˅ q) ˄ ∼p
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Following is the probability distribution of a r.v.X.
| X | – 3 | – 2 | –1 | 0 | 1 | 2 | 3 |
| P(X = x) | 0.05 | 0.1 | 0.15 | 0.20 | 0.25 | 0.15 | 0.1 |
Find the probability that X is positive.