Suppose that 6% of people are left handed and suppose you pick two people at random. Rounding your answer to two decimal places, find the probability that

The probability that neither person is left-handed is 88.36%, that only one is left-handed is 5.64%, and that both are left-handed is 0.36%.

Step-by-step explanation:

Supposing that 6% of people are left handed and supposing two people are picked at random, to find the probability that 0, 1 or 2 people are left-handed, the following calculation must be performed:

100 - 6 = 94

0 left-handed people = 0.94 x 0.94 = 0.8836 = 88.36%

1 left-handed person = 0.94 x 0.06 = 0.0564 = 5.64%

2 left-handed people = 0.06 x 0.06 = 0.0036 = 0.36%

Therefore, the probability that neither person is left-handed is 88.36%, that only one is left-handed is 5.64%, and that both are left-handed is 0.36%.

b.my answer is b.

brainliest

any options?

step-by-step explanation: