[amp_mcq option1=”$${\\left( {\\frac{1}{2}} \\right)^{{\\text{N}} – 1}}$$” option2=”$$1 – {\\left( {\\frac{1}{2}} \\right)^{{\\text{N}} – 1}}$$” option3=”$${\\left( {\\frac{1}{2}} \\right)^{\\text{N}}}$$” option4=”$$1 – {\\left( {\\frac{1}{2}} \\right)^{\\text{N}}}$$” correct=”option1″]<\/p>\n
The correct answer is $\\boxed{B. 1 – {\\left( {\\frac{1}{2}} \\right)^{{\\text{N}} – 1}}}$.<\/p>\n
A fair coin has two sides, heads and tails. When a fair coin is tossed, the probability of getting heads is $\\frac{1}{2}$ and the probability of getting tails is also $\\frac{1}{2}$.<\/p>\n
If a fair coin is tossed $N$ times, the probability of getting heads all $N$ times is $\\left( {\\frac{1}{2}} \\right)^{{\\text{N}}}$.<\/p>\n
The probability of not getting heads in any of the tosses is the complement of the probability of getting heads all $N$ times. The complement of an event is the event that does not occur. So, the probability of not getting heads in any of the tosses is $1 – \\left( {\\frac{1}{2}} \\right)^{{\\text{N}}}$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”$${\\left( {\\frac{1}{2}} \\right)^{{\\text{N}} – 1}}$$” option2=”$$1 – {\\left( {\\frac{1}{2}} \\right)^{{\\text{N}} – 1}}$$” option3=”$${\\left( {\\frac{1}{2}} \\right)^{\\text{N}}}$$” option4=”$$1 – {\\left( {\\frac{1}{2}} \\right)^{\\text{N}}}$$” correct=”option1″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[691],"tags":[],"class_list":["post-20308","post","type-post","status-publish","format-standard","hentry","category-probability-and-statistics","no-featured-image-padding"],"yoast_head":"\n