Check whether the following probabilities $\mathrm{P}(\mathrm{A})$ and $\mathrm{P}(\mathrm{B})$ are consistently defined
(i) $\mathrm{P}(\mathrm{A})=0.5, \mathrm{P}(\mathrm{B})=0.7, \mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.6$
(ii) $\mathrm{P}(\mathrm{A})=0.5, \mathrm{P}(\mathrm{B})=0.4, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.8$

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Video Transcript

in this problem, let us first look at the first part, We have a plurality of S .5. Quality of b. .7 and intersection years .6. Now we know that if A. And B are defined, then be a intersection B has to be less than the probability of E. But in this case Be intersection sees .6. Let's stick whether it is less than 2.5, which means that does not hold. So since being intersection B is not less than B. Of A. That means P. A. And P. B are not consistently defined. So here baby, not consistently design. Let's look at the second part, which is pia is 0.5. Bbs 0.4 and be a union B. S. Point it. Now we know that the union be has to be greater than being and be a union. Be has to be written than baby. If you check in this case, then The intersection which is pointed is greater than both .5 and .4. So this holds, which means that B. And B. B. Uh consistently you're fine. Not so

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