So lets see Quantity A: Both get degree = \(P(tom) \times P(john)=0.5 *0.4=0.2\).

Quantity B: Atleast one degree gets awarded. We can write the following cases:

Tom gets a degree John doesnot = \(0.5 * (1-0.4)=0.3\)

John gets a degree Tom doesnot = \(0.4 * (1-0.5)=0.2\)

Both get a degree = \(0.4 *0.5=0.2\)

Total probability for quantity B is \(0.7\).

Hence B is greater.
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I do not think we have to do serious math to answer this question. Both events happening vs any one event happening. B would be greater for obvious reason.

Both events happening = 0.5*0.4 = 0.2

any one event happening = 0.5+0.4 - 0.9

Please correct me on Math part if wrong.

I could figure out the A answer but I was not so sure of the method for B. Help please?

Proability that Tom passes the test is 0.5, So, probability that he does not pass the test is also 0.5. and the probability that John passes the test is 0.4. So, the probability that John doesn't pass the test is 0.6. (1-P) equation.

So, probability of both get degree is 0.5*0.4= 0.2.

While probability of both don't get degree is 0.5*0.6 =0.3
Hence, probability of atleast one of them gets degree is (1-0.3) 0.7

Ans. B

Answer: B
P(Tom passes) = 0.5
So, P(Tom Fails) = 1 - 0.5 = 0.5

P(John passes) = 0.4
So, P(Tom Fails) = 1 - 0.4 = 0.6

The two events are independent from each other.

A: P(Tom passes and John passes) = P(Tom passes) * P(John passes) = 0.5 * 0.4 = 0.2 [because events are independent, their intersection probability is multiplication of the individuals)

B: P(Tom passes and John passes) + P(Tom passes and John fails) + P(Tom fails and John passes) = 1 - P(Tom fails and John fails) = 1 - P(Tom fails)*P(John fails) = 1 - 0.5*0.6 = 1 - 0.3 = 0.7
So B is bigger than A.
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Both of them getting the degree is 0.50*.40 = .20
The chance of either getting the degree will be P(A) + P(B) - P(AB). Which is 0.70

So clearly B is greater.

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