Ajay vs. Sanjay

Probability Level 2

Archery players Ajay and Sanjay can hit the bulls-eye with accuracy rates of $0.6$ and $0.8$ , respectively. If each person shoots at the target once, what is the probability that either Ajay or Sanjay (or both) hits the bulls-eye?

0.7

0.92

0.88 0.9

3 solutions

Sunit Bhattacharya
May 5, 2014

Probability of Ajay hitting = 0.6 and that of missing = 0.4 Probability of Sanjay hitting = 0.8 and that of missing = 0.2 So required probability is-: P(Ajay Missing and Sanjay hitting) or P(Ajay Hitting and Sanjay Missing) orP(Ajay and Sanjay both hitting) =(0.6 0.2)+(0.8 0.4)+(0.6*0.8) =0.12+0.32+0.48 =0.92

Pavan Kumar Vaitheeswaran
Mar 12, 2014

The event of either of them scoring is complementary to that when both of them do not. Hence, required probability, P = 1- P(Ajay misses) $\times$ P(Sanjay misses) = 1-(1-P(Ajay scores)) $\times$ (1-P(Sanjay scores)) = $1-0.4 \times 0.2 = 0.92$

the actual answer must me (0.8 $times$ 0.4)+(0.2 \times 0.8)= 0.44 because in the problem it is stated that either ajay hit or sanjay hit that doesn't mean both can hit at same instance, and to get the answer as 0.92 we have to take the case in which both hits the target which will add (0.8 \times 0.6)=0.48 to my answer(0.44) and will also get answer as given

Kunal Mandil - 7 years, 1 month ago

It says in the problem that it either one or the other, or both., therefore we are looking for the probability of (ajaybullseye U sanjaybullseye) : One, or the other, or both.

Nicolas Bryenton - 7 years, 1 month ago

Pragya Singhal
May 6, 2014

STEP 1 :- P(Either Ajay or Sanjay) = P(Ajay Hit) + P(Sanjay Hit) - P( Ajay and Sanjay Hit) STEP 2 :- P(Either Ajay or Sanjay) = 0.6+0.8-(0.6x0.8) =0.92

Ajay vs. Sanjay

3 solutions

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