Advertisements
Advertisements
प्रश्न
Solve the following question and mark the best possible option.
Let x, y, z are 3 numbers (not necessarily distinct) chosen randomly with replacement. Also, x, y, z ∈ {1,2,3,4,5}. What is the probability that xy + z is even?
विकल्प
`2/5`
`1/2`
`64/125`
`59/125`
MCQ
उत्तर
2 cases arise:-
Case-1: xy & z both are odd
⇒ x, y, z all are odd
⇒ x, y, z = 1 or 3 or 5
So, probability = `3/5 xx 3/5 xx 3/5 = 27/125`
Case-2: Both xy & z are even.
Probability of z being even = `2/5`
Probability of xy being even = 1 - P(xy being odd)
= 1- `3/5 xx 3/5 = 16/25`.
So, required probability is `27/125 + 2/5 × 16/25= (27+32)/125 = 59/125`.
shaalaa.com
Probability (Entrance Exam)
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?